Content-Type: multipart/mixed; boundary="-------------9904151203928" This is a multi-part message in MIME format. ---------------9904151203928 Content-Type: text/plain; name="99-116.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="99-116.keywords" Reaction Diffusion Equations Asymptotic Analysis Matched Expansions ---------------9904151203928 Content-Type: application/x-tex; name="final006.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="final006.tex" %% This document created by Scientific Word (R) Version 3.0 \documentclass[a4paper,notitlepage]{article} \usepackage{amssymb} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{latexsym} \usepackage{amsmath} \usepackage{graphicx} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Created=Thu Oct 15 19:16:28 1998} %TCIDATA{LastRevised=Thu Apr 15 14:31:51 1999} %TCIDATA{} %TCIDATA{} %TCIDATA{Language=American English} %TCIDATA{CSTFile=LaTeX article.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{Counters=arabic,1} \textheight267mm \textwidth190mm \topmargin-25.4mm \oddsidemargin-15.4mm \evensidemargin-15.4mm \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \title{Asymptotics of solutions in $nA+nB\rightarrow C$ reaction--diffusion systems} \author{Guillaume van Baalen\thanks{% Supported in part by the Fonds National Suisse.} \\ %EndAName D\'{e}partement de Physique\\ Th\'{e}orique\\ Universit\'{e} de Gen\`{e}ve\\ Switzerland\\ vanbaal4@kalymnos.unige.ch \and Alain Schenkel\thanks{% Supported by the Fonds National Suisse and DOE grant SCUOCB341495.} \\ %EndAName Helsinki Institute of Physics\\ University of Helsinki\\ Finland\\ alain.schenkel@helsinki.fi \and Peter Wittwer\thanks{% Supported in part by the Fonds National Suisse.} \\ %EndAName D\'{e}partement de Physique\\ Th\'{e}orique\\ Universit\'{e} de Gen\`{e}ve\\ Switzerland\\ wittwer@ibm.unige.ch} \maketitle \begin{abstract} We analyze the long time behavior of initial value problems that model a process where particles of type $A$ and $B$ diffuse in some substratum and react according to $nA+nB\rightarrow C.$ The case $n=1$ has been studied before; it presents nontrivial behavior on the reactive scale only. In this paper we discuss in detail the cases $n>3,$ and prove that they show nontrivial behavior on the reactive and the diffusive length scale.\newpage \end{abstract} \tableofcontents \section{Introduction and main results} We consider, for arbitrary but fixed $n\in\mathbf{N,}$ $n\geq1,$ the reaction--diffusion problem \begin{align} a_{t} & =a_{xx}-\frac{1}{2}(4ab)^{n}~, \label{system1} \\ b_{t} & =b_{xx}-\frac{1}{2}(4ab)^{n}~, \label{system2} \end{align} for $x\in\mathbf{R},$ $t\geq\tau\geq0,$ with initial conditions $a(x,\tau )=a_{0}(x),$ $b(x,\tau)=b_{0}(x),$ satisfying \begin{align} \lim_{x\rightarrow-\infty}a_{0}(x) & =1~, \notag \\ \lim_{x\rightarrow+\infty}b_{0}(x) & =1~, \label{lim002} \\ \lim_{x\rightarrow+\infty}a_{0}(x) & =\lim_{x\rightarrow-\infty}b_{0}(x)=0~. \notag \end{align} The choice of the initial time $t=\tau,$ and a class of initial conditions $% a_{0},$ $b_{0}$ will be described later on, but for the purpose of this introduction it is useful to have in mind the ``natural'' case: $\tau=0,$ $% a_{0}(x)=1$ for $x<0,$ $a_{0}(x)=0$ for $x>0,$ and $b_{0}(x)=1$ for $x>0,$ $% b_{0}(x)=0$ for $x<0.$ This initial value problem models the time evolution of a chemical system of two (initially separated) substances $A$ and $B,$ that diffuse in some substratum and react according to $nA+nB\rightarrow C,$ with a substance $C$ that is supposed not to participate in the reaction anymore. The model is a mean--field description of such a situation where the functions $a$ and $b$ represent the densities of the substances $A$ and $B.$ For more details see \cite{Droz}. Equations (\ref{system1}) and (\ref{system2}) are best studied in terms of the sum \begin{equation} v=a+b~, \label{defv} \end{equation} and the difference \begin{equation} u=a-b~, \label{defu} \end{equation} which satisfy the equations \begin{align} u_{t} & =u_{xx}~, \label{equ} \\ v_{t} & =v_{xx}-(v^{2}-u^{2})^{n}~, \label{eqv} \end{align} with initial conditions $v_{0}$ and $u_{0}$ (at time $t=\tau)$ that satisfy \begin{align} \lim_{x\rightarrow-\infty}u_{0}(x) & =1~, \notag \\ \lim_{x\rightarrow+\infty}u_{0}(x) & =-1~, \label{limitu} \end{align} and \begin{equation} \lim_{x\rightarrow\pm\infty}v_{0}(x)=1~. \label{limitv} \end{equation} For initial conditions $a_{0},$ $b_{0}$ with \begin{equation} a_{0}(x)=b_{0}(-x)~, \label{sym001} \end{equation} the functions $v_{0}$ and $u_{0}$ are even and odd, respectively, and the equations (\ref{equ}), (\ref{eqv}) preserve this symmetry. Furthermore, for the special initial condition \begin{equation} u(x,\tau)=-\mu_{1}(x/\sqrt{\tau})~, \label{initsym} \end{equation} with $\mu_{1}$ defined by the equation \begin{equation} \mu_{1}(y)=\mathrm{erf}(\frac{y}{2})\equiv\frac{2}{\sqrt{\pi}}% \int_{0}^{y/2}e^{-\sigma^{2}}d\sigma~, \label{defmu} \end{equation} equation (\ref{equ}) has the explicit solution \begin{equation} u(x,t)=-\mu_{1}(x/\sqrt{t})~. \label{uxt} \end{equation} We note that the initial condition (\ref{initsym}) for $u$ (at time $t=\tau$% ) is simply the solution of equation (\ref{equ}) with the ``natural'' initial condition, $u(x,0)=1$ for $x<0,$ $u(x,0)=-1$ for $x>0,$ evaluated at $t=\tau.$ To keep this paper as simple as possible we now restrict the discussion to this case, i.e., we consider from now on equation (\ref{eqv}) with initial conditions satisfying (\ref{limitv}), and $u$ given by (\ref {uxt}). We note, however, that more general (asymmetric) initial conditions for $u$ could be treated as well. This would lead to corrections to $u$ of the order $\mathcal{O}(1/t),$ and such corrections do not change in any way the discussion of the equation for $v$ that follows. The reaction--diffusion problems considered here develop, in addition to the built--in diffusive length scale $\mathcal{O}(\sqrt{t}),$ an additional shorter length scale, on which the reaction takes place. The function $F,$% \begin{equation} F=\frac{1}{2}(4ab)^{n}\equiv \frac{1}{2}(v^{2}-u^{2})^{n}~, \label{reaction001} \end{equation} is called the reaction term or reaction front, and we are interested in describing the asymptotic behavior of the function $F$ for large times. The knowledge of this behavior is useful, since it appears to be universal, in the sense that it is largely independent of the choice of the initial conditions and of the details of the model under consideration. As mentioned above, if $v_{0}$ is an even function, then $v$ and as a consequence $F$ are even functions of $x.$ We will see that the critical point of $F$ at $x=0$ is a maximum, and that $F$ decays (rapidly) for large $x.$ Before proceeding any further we note that the factor of $4^{n-1/2}$ in (\ref {system1}), (\ref{system2}) and (\ref{reaction001}) is just a normalization, and has been chosen for convenience to make the equation (\ref{eqv}) for $v$ look simple. In fact, any system of the form \begin{align*} a_{t} & =D_{a}a_{xx}-k_{a}(ab)^{n}~, \\ b_{t} & =D_{b}b_{xx}-k_{b}(ab)^{n}~, \end{align*} with positive $D_{a},$ $D_{b},$ $k_{a},$ and $k_{b},$ and with initial conditions such that \begin{align*} \lim_{x\rightarrow-\infty}a(x,0) & =a_{\infty}>0~, \\ \lim_{x\rightarrow\infty}b(x,0) & =b_{\infty}>0~, \\ \lim_{x\rightarrow+\infty}a(x,0) & =\lim_{x\rightarrow-\infty}b(x,0)=0~, \end{align*} can be reduced, by scaling space and time and the amplitudes, to the problem \begin{align*} a_{t} & =a_{xx}-\frac{1}{2}(4ab)^{n}~, \\ b_{t} & =Db_{xx}-\frac{1}{2}(4ab)^{n}~, \end{align*} with $D>0,$ and with initial conditions such that \begin{align*} \lim_{x\rightarrow-\infty}a(x,0) & =1~, \\ \lim_{x\rightarrow\infty}b(x,0) & =\beta>0~, \\ \lim_{x\rightarrow+\infty}a(x,0) & =\lim_{x\rightarrow-\infty}b(x,0)=0~. \end{align*} In this paper we have limited the discussion to the case $\beta=1$ and $D=1. $ The case $\beta\neq1$ leads to a moving reaction front. A change of coordinates to a moving frame complicates the analysis, but the problem could still be treated with the methods presented here. Choosing $D=1$ makes the equations mathematically simpler. As a consequence, as we have seen, the two equations for $a$ and $b$ can be reduced to just one equation for the sum $v=a+b,$ since the equation for the difference $u=a-b$ can be solved explicitly. Even though we do not expect the asymptotic behavior of the solution to change in any relevant way if $D\neq1$, the strategy of proof would have to be changed considerably, since the equations can not be decoupled anymore in that case. Before we state our results, we briefly discuss the expected dependence of the results on the parameter $n.$ The case $n=1$ has been studied in detail in \cite{Wittwer}, where it is proved that in this case the reaction term (\ref{reaction001}) satisfies, for all $z\in\mathbf{R},$ \begin{equation*} \lim_{t\rightarrow\infty}t^{2\gamma}F(t^{\alpha}z,t)=\rho(\left| z\right| )~, \end{equation*} where $\alpha=1/6,$ and $\gamma=1/3,$ and where $\rho\colon\mathbf{R}_{+}% \mathbf{\rightarrow R}_{+}$ is a smooth function that decays like $\exp(-% \mathrm{const.}z^{3/2})$ for large values of $z.$ It follows furthermore from the results in \cite{Wittwer} that the function $F$ is very small on the diffusive scale in the sense that for $n=1,$ $y\neq0,$ and all $p\geq0, $% \begin{equation} \lim_{t\rightarrow\infty}t^{p}F(\sqrt{t}y,t)=0~. \label{fdiffuse} \end{equation} The smallness of $F$ on the diffusive scale is easily understood by realizing that, for $n=1$ and for positive values of $x$ on the diffusive scale, i.e., for $x/\sqrt{t}>>1,$ equation (\ref{system1}) essentially reduces to \begin{equation} a_{t}=a_{xx}-\lambda a~, \label{a001} \end{equation} with $\lambda>0.$ Therefore, the function $a$ decays exponentially fast to zero on this scale, and similarly for $b$ for negative values of $x.$ For $n>1,$ however, equation (\ref{system1}) reduces, for $x/\sqrt{t}>>1,$ essentially to \begin{equation} a_{t}=a_{xx}-\lambda a^{n}~, \label{a002} \end{equation} with $\lambda>0.$ The solution of (\ref{a002}) has an asymptotic behavior that is radically different from the solution of (\ref{a001}). In particular, for $n=2$, the solution may even blow up in finite time if $a$ is not a positive function. Note that, for $n$ odd, the nonlinear term in (% \ref{a002}) is always a ``friction term'', independent of the sign of $a,$ and the case of $n$ odd will therefore turn out to be easier to treat than the case of $n$ even. It is well known \cite{kupi} that for $n>3$ and small bounded integrable initial conditions, the nonlinearity in (\ref{a002}) becomes irrelevant for large times in the sense that the solution converges to a multiple of $\exp (-x^{2}/4t)/\sqrt{t},$ which solves the linear equation $a_{t}=a_{xx}.$ We would therefore expect that, for $n>3,$ the function $F$ is of the order $\mathcal{O}(t^{-n/2})$ on the diffusive scale. This turns out to be wrong. As we will prove below, $F$ is of the order $% \mathcal{O}(t^{-n/(n-1)})$ for $n>3,$ because $F$ converges on this scale pointwise to a function that is not integrable at the origin. This corresponds to a solution of (\ref{a002}) for which the nonlinear term is a marginal perturbation, i.e., a solution with an amplitude of the order $% \mathcal{O}(t^{-1/(n-1)})$. We will see that one can take advantage of this fact, and a diffusive stability bound will be good enough to prove convergence of $F$ to its limit, but as a consequence, our results will be limited to the case $n>3.$ The cases $n=2$ and $n=3$ are special and will not be discussed any further. The following theorem is our main result. \begin{theorem} \label{main}For arbitrary but fixed $n\in\mathbf{N,}$ $n\geq4,$ there exist $% \tau>0,$ functions $\mu_{1,}$ $\mu_{2},$ $\varphi_{1},$ $\varphi_{2},$ and a class of initial conditions (specified at $t=\tau),$ such that (\ref{eqv}) has a unique solution $v$ that satisfies for all $t\geq\tau$ the bound \begin{equation} \left| v(x,t)-v_{\infty}(x,t)\right| <\frac{\mathrm{const.}}{t^{4\gamma}}~, \label{bound} \end{equation} where \begin{equation} v_{\infty}(x,t)=\mu_{1}(\frac{\left| x\right| }{\sqrt{t}})+t^{-\varepsilon }\mu_{2}(\frac{\left| x\right| }{\sqrt{t}})+t^{-\gamma}\varphi_{1}(\frac{% \left| x\right| }{t^{\alpha}})+t^{-3\gamma}\varphi_{2}(\frac{\left| x\right| }{t^{\alpha}})~, \label{vinfty} \end{equation} $\gamma=\frac{1}{2n+1},$ $\varepsilon=\frac{1}{n-1}$ and $\alpha=\frac{1}{2}% -\gamma.$ \end{theorem} \begin{remark} This theorem is a local result, in the sense that the class of initial conditions will be a set of functions in a (small) neighborhood of the function $v_{\infty,0},$ $v_{\infty,0}(x)=v_{\infty}(x,\tau).$ In particular, our methods do not allow us to show that the solution with the ``natural'' initial condition $v_{0}\equiv1$ at $t=0$ belongs to this set at $t=\tau.$ We do expect, however, that this is the case, as has been proved for $n=1$ in \cite{Wittwer}. \end{remark} \begin{remark} We note that, if an initial condition $v_{0}$ is such that $v_{0}(x)-\left| u(x/\sqrt{\tau})\right| <0$ for a certain $x,$ then $a_{0}(x)<0$, if $x>0,$ or $b_{0}(x)<0$ if $x<0.$ A priori, we do not need to consider such initial conditions, since in our model $a$ and $b$ represent particle densities, and the solutions $a$ and $b$ are positive if the initial conditions $a_{0}$ and $b_{0}$ are positive. As we will see, for $n\geq4,$ it will not be necessary to impose that $a_{0}$ and $b_{0}$ be positive everywhere, and it will neither be necessary to impose that $v_{0}=a_{0}+b_{0}$ be an even function. \end{remark} As we will see, the functions $\varphi _{1}$ and $\varphi _{2}$ are small on the diffusive scale, i.e., for $x\approx \sqrt{t}y,$ $y\neq 0,$ and $t$ large, \begin{equation} v_{\infty }(\sqrt{t}y,t)=\mu _{1}(\left| y\right| )+t^{-\varepsilon }\mu _{2}(\left| y\right| )+\mathcal{O}(t^{-2\varepsilon ^{\prime }})~, \label{asym001} \end{equation} where $\varepsilon ^{\prime }=\varepsilon $ if $n>5,$ and $2\gamma <\varepsilon ^{\prime }<\varepsilon $ if $n=4,$ $5.$ Using the definition (% \ref{defv}), (\ref{defu}) for $v$ and $u,$ we therefore find that for $y>0$ and $t$ large, \begin{equation*} a(\sqrt{t}y,t)=\frac{1}{2}t^{-\varepsilon }\mu _{2}(y)+\mathcal{O}% (t^{-2\varepsilon ^{\prime }})~, \end{equation*} and similarly for $b,$ for $y<0.$ In contrast to the case $n=1,$ where only exponentially few particles reach the diffusive scale, the amount of particles decays only slowly for $n>3.$ Our results imply that, for large times, the density of the remaining particles is given by the function $\mu _{2},$ i.e., it is independent of the initial conditions. As a corollary to Theorem \ref{main} we get a precise description of the reaction front $F$ on the reactive and the diffusive scale. This description will be given in Section 4, once we have defined the functions $\mu _{1},$ $% \mu _{2},$ $\varphi _{1}$ and $\varphi _{2}$ in Section 3. In Section 2 we explain our strategy for proving Theorem \ref{main}. This strategy is implemented in Section 5 and Section 6. The Appendix contains the proof of the existence of the functions $\mu _{2},$ $\varphi _{1}$ and $\varphi _{2}$. \section{Strategy of the Proof} Consider functions $v$ of the form \begin{equation} v(x,t)=v_{\infty}(x,t)+\psi(x,t)~, \label{ansatz} \end{equation} with $v_{\infty}$ as in Theorem \ref{main}, and $\psi(x,\tau)=\psi_{0}(x),$ for some $\tau>>1,$ with $\psi_{0}\in L_{1}\cap L_{\infty}.$ Substituting (% \ref{ansatz}) into (\ref{eqv}) leads to an equation for the function $\psi$ of the form \begin{equation} \dot{\psi}=\psi^{\prime\prime}-V\psi-I-T(\psi)~, \label{eqpsi} \end{equation} for certain functions $V$ and $I,$ and for $T$ some nonlinear map. We will show that if $v_{\infty}$ is defined correctly, $\tau$ large enough and $% \psi_{0}$ small enough, then $V$ can be chosen positive and $T$ will be small, so that the solution of equation (\ref{eqpsi}) will be bounded for large times by the corresponding solution of the inhomogeneous heat equation $\dot{\psi}=\psi^{\prime\prime}-I.$ We will find that, with the right choice of $v_{\infty},$% \begin{equation} \int dx~\left| I(\sqrt{t}x,t)\right| \leq\mathrm{const.}\text{ }% t^{-1-4\gamma}~, \label{boundi} \end{equation} from which the bound (\ref{bound}) will follow. We note that $4\gamma<\frac {1}{2}$ for $n\geq4>\frac{7}{2},$ so that contributions of initial conditions will become irrelevant for large times, i.e., the solution $\psi$ becomes what is called ``slaved to the inhomogeneous term''. \section{Asymptotic Expansion} In order to implement the strategy outlined in Section 2, we need a function $v_{\infty }$ that approximates the solution $v$ for large times sufficiently well, uniformly in $x.$ Since we would like to control the time evolution of equation (\ref{eqpsi}) on $L_{1}\cap L_{\infty },$ this function $v_{\infty }$ needs to satisfy $\lim_{x\rightarrow \pm \infty }v_{\infty }(x,t)=1$ in order for $v$ to satisfy the boundary conditions (% \ref{limitv}). Furthermore, the inhomogeneous term $I$ in equation (\ref {eqpsi}) contains second derivatives of $v_{\infty },$ and the function $I$ can therefore only be in $L_{1}\cap L_{\infty }$ if $v_{\infty }$ is at least twice differentiable. We now construct a function $v_{\infty }$ satisfying these requirements through a two length--scale asymptotic expansion. To simplify the notation later on we use the convention that, unless stated otherwise, \begin{equation} y\equiv\frac{x}{\sqrt{t}}~, \label{defy} \end{equation} and \begin{equation} z\equiv\frac{x}{t^{\alpha}}~, \label{defz} \end{equation} and we will refer to $y$ as the diffusive length scale and to $z$ as the reactive length scale. The function $v_{\infty}$ is given by the first and second order terms of a so called ``matched asymptotic expansion''. The ``matched'' refers to the fact that such an expansion contains functions that can not be associated uniquely with one of the length scales and can therefore be used to ``match'' the behavior at large distances of the shorter scale with the behavior at short distances of the larger scale. Let \begin{equation} \delta=\frac{n+2}{n-1}~, \label{defdelta007} \end{equation} and let $\gamma,$ $\alpha,$ and $\varepsilon$ be as in Theorem \ref{main}. Then, the functions $y\equiv t^{-\gamma}z,$ $t^{-\varepsilon}y^{-\delta}% \equiv t^{-\gamma}z^{-\delta},$ $t^{-\varepsilon}y^{2-\delta}\equiv t^{-3\gamma }z^{2-\delta}$ and $t^{-2\varepsilon}y^{-2\delta+1}\equiv t^{-3\gamma }z^{-2\delta+1}$ are of this form and will naturally show up in the function $v_{\infty}.$ As a consequence, the representation (\ref{vinfty}% ) for $v_{\infty}$ is not unique. If we choose (as we will) to compute the expansion for $v_{\infty}$ in the order of decreasing amplitudes, i.e., if we first compute the term of order $\mathcal{O}(t^{-\gamma}),$ then the term of order $\mathcal{O}(t^{-\varepsilon}),$ and finally the term of order $% \mathcal{O}(t^{-3\gamma}),$ we get a representation of $v_{\infty}$ of the form \begin{equation} v_{\infty}(x,t)=\mu_{1}(\left| y\right| )+t^{-\gamma}\eta(\left| z\right| )+t^{-\varepsilon}\left( \mu_{2}(\left| y\right| )-\lambda\left| y\right| ^{-\delta}\right) +t^{-3\gamma}\varphi_{2}(\left| z\right| )~, \label{repres1} \end{equation} where \begin{equation} \eta(z)=\varphi_{1}(z)+\lambda z^{-\delta}~, \label{etaphi1} \end{equation} with $\lambda$ a certain constant to be determined later. We note that, by definition, $v_{\infty}$ is a symmetric function, and it is therefore sufficient to consider positive values of $x$ if we choose appropriate boundary conditions at $x=0$ to ensure regularity. Finally, since we will need to describe the asymptotic behavior of various functions near zero and infinity, we introduce the following notation. Let $% f $ be a continuous function from $\mathbf{R}_{+}$ to $\mathbf{R,}$ $k$ a positive integer and $p_{1}0~, \label{lambda0} \end{equation} with $\lambda _{1}\neq 0$ and with $\lambda $ as in Proposition \ref {theoremeta}. For $y$ large, the function $\mu _{2}$ decays rapidly in the sense that \begin{equation} \mu _{2}(y)=\exp (-\frac{y^{2}}{4})\left( \frac{C}{y^{1-2\varepsilon }}% +\dots \right) ~, \label{asymmu2} \end{equation} for some constant $C>0.$ \end{proposition} \subsection{Equation for $\protect\varphi_{2}$} We make the ansatz $v(x,t)=\mu_{1}(\left| y\right| )+t^{-\gamma}\eta(\left| z\right| )+t^{-\varepsilon}(\mu_{2}(\left| y\right| )-\lambda\left| y\right| ^{-\delta})+t^{-3\gamma}\varphi_{2}(\left| z\right| )$ which we substitute into equation (\ref{eqv}). We multiply the resulting equation with $% t^{3\gamma+2\alpha},$ and take then the limit $t\rightarrow\infty,$ keeping $% z$ fixed. This leads to the (linear) differential equation for $\varphi _{2}, $ \begin{equation} \varphi_{2}^{\prime\prime}+\gamma\eta+\alpha z\eta^{\prime}+(2-\delta )(1-\delta)\lambda_{0}z^{-\delta}=n\left( 2\kappa z\eta+\eta^{2}\right) ^{n-1}\left[ (2\kappa z+2\eta)(\varphi_{2}+\lambda_{0}z^{2-\delta})+2\kappa_{3}z^{3}\eta\right] ~. \label{varphi2} \end{equation} In order to compensate the sub--leading singular behavior of $\mu_{2}$ near $% x=0$ we make the ansatz \begin{equation} \varphi_{2}(z)=-\lambda_{0}z^{2-\delta}+h(z)~, \label{newh} \end{equation} which we substitute into equation (\ref{varphi2}). For the function $h$ we get the equation \begin{equation} h^{\prime\prime}+\gamma\eta+\alpha z\eta^{\prime}=n\left( 2\kappa z\eta +\eta^{2}\right) ^{n-1}\left( (2\kappa z+2\eta)h+2\kappa_{3}z^{3}\eta\right) ~. \label{h} \end{equation} Since the function $\eta$ is regular near $z=0,$ the solution $h$ turns out to be regular near $z=0,$ too. Therefore, the function $z\mapsto h(\left| z\right| )$ is twice differentiable near $x=0$ if we impose at $z=0$ the boundary condition \begin{equation} h^{\prime}(0)=0~. \label{bh-1} \end{equation} At infinity we need that $\lim_{z\rightarrow\infty}\varphi_{2}(z)=0.$ We therefore require that \begin{equation} \lim_{z\rightarrow\infty}(h(z)-\lambda_{0}z^{2-\delta})=0~. \label{bh-2} \end{equation} A proof of the following proposition can be found in the appendix. \begin{proposition} \label{cfi22}For all $n\geq 4,$ there exists a unique function $h\colon \mathbf{R}_{+}\rightarrow \mathbf{R}$ that satisfies equation (\ref{h}) with the boundary conditions (\ref{bh-1}) and (\ref{bh-2}). Near $z=0,$ the function $h$ is of the form \begin{equation*} h(z)=h_{0}+h_{2}z^{2}+\dots ~, \end{equation*} with certain coefficients $h_{0}$ and $h_{2},$ and for $z$ large $h$ is of the form \begin{equation*} h(z)=\lambda _{0}z^{2-\delta }+\frac{\lambda ^{\prime }}{z^{\delta ^{\prime }-2}}+\dots ~, \end{equation*} with $\lambda _{0}$ as defined in (\ref{lambda0}), for a certain constant $% \lambda ^{\prime }$, and with $\delta ^{\prime }$ as defined in Proposition \ref{theoremeta}. \end{proposition} \section{The Reaction Front} Using the properties of the functions $\mu_{1},$ $\mu_{2},$ $\varphi_{1}$ and $\varphi_{2},$ we get from Theorem \ref{main} the following behavior of the reaction front $F.$ \begin{corollary} Let $v$ be as in Theorem \ref{main}, and $F$ as defined in (\ref{reaction001}% ). Then, for all $z\in \mathbf{R,}$% \begin{equation*} \lim_{t\rightarrow \infty }t^{2n\gamma }F(t^{\alpha }z,t)=\frac{1}{2}% (2\kappa \left| z\right| \eta (\left| z\right| )+\eta (\left| z\right| )^{2})^{n}=\frac{1}{2}\eta ^{\prime \prime }(\left| z\right| )=\left\{ \begin{array}{lll} \eta _{2}-6\eta _{4}\left| z\right| ^{2}+\dots & \text{for} & \noindent \left| z\right| \approx 0~, \\ & & \\ \frac{1}{2}\left( 2\lambda \kappa \right) ^{n}/\left| z\right| ^{\delta +2}+\dots & \text{for} & \noindent \left| z\right| >>1~, \end{array} \right. \end{equation*} and for all $y\neq 0,$% \begin{equation*} \lim_{t\rightarrow \infty }t^{n\varepsilon }F(\sqrt{t}y,t)=\frac{1}{2}(2\mu _{1}~\mu _{2})^{n}(\left| y\right| )=\left\{ \begin{array}{lll} \frac{1}{2}\left( 2\lambda \kappa \right) ^{n}/\left| y\right| ^{\delta +2}+\dots & \text{for} & \left| y\right| \approx 0~, \\ & & \\ \exp (-n\left| y\right| ^{2}/4)(2^{n-1}C^{n}/\left| y\right| ^{n(1-2\varepsilon )}+\dots ) & \text{for} & \left| y\right| >>1~. \end{array} \right. \end{equation*} Here, $\eta _{2},$ $\eta _{4}$ are as defined in Proposition \ref{theoremeta} and $C$ is as defined in (\ref{asymmu2}). \end{corollary} \section{The Equation for $\protect\psi$} In order to simplify the notation we define the function $\overline{u},$ \begin{equation*} \overline{u}(x,t)=\mu_{1}(\left| y\right| )~, \end{equation*} the function $\mu_{3},$% \begin{equation*} \mu_{3}(y)=\mu_{2}(y)-\lambda y^{-\delta}~, \end{equation*} the function $\phi,$% \begin{equation} \phi(x,t)=t^{-\gamma}\eta(|z|)+t^{-\varepsilon}\mu_{3}(|y|)+t^{-3\gamma }\varphi_{2}(|z|)~, \label{defphi} \end{equation} and the function $\phi_{1},$% \begin{equation} \phi_{1}(x,t)=\phi(x,t)-\kappa\frac{\left| x\right| }{\sqrt{t}}~. \label{defphi1} \end{equation} The function $v_{\infty}$ in Theorem \ref{main} and in (\ref{repres1}) can then be written as $v_{\infty}=\overline{u}+\phi.$ Let now $v=v_{\infty}+\psi.$ Then, \begin{align*} \left( v^{2}-u^{2}\right) ^{n} & =\left( \left( \overline{u}+\phi +\psi\right) ^{2}-u^{2}\right) ^{n}=\left( \left( 2\overline{u}\phi +\phi^{2}\right) +\left( 2(\overline{u}+\phi)+\psi\right) \psi\right) ^{n} \\ & =\sum_{k=0}^{n}\binom{n}{k}\left( 2\overline{u}\phi+\phi^{2}\right) ^{n-k}\left( 2(\overline{u}+\phi)+\psi\right) ^{k}\psi^{k}~. \end{align*} Therefore, substituting the ansatz $v=v_{\infty}+\psi$ into (\ref{eqv}) leads to the following equation for the function $\psi,$% \begin{equation} \dot{\psi}=\psi^{\prime\prime}-\widehat{V}\psi-I-\widehat{T}(\psi )~, \label{equpsi1} \end{equation} with the function $\widehat{V},$ \begin{equation} \widehat{V}=2n~(2\overline{u}\phi+\phi^{2})^{n-1}~(\overline{u}+\phi )~, \label{vhat} \end{equation} the function $I,$% \begin{equation} I=-\dot{\phi}+\phi_{1}^{\prime\prime}-\left( 2\overline{u}\phi+\phi ^{2}\right) ^{n}~, \label{equi} \end{equation} and the map $\widehat{T},$% \begin{equation} \widehat{T}(\psi)=n\left( 2\overline{u}\phi+\phi^{2}\right) ^{n-1}\psi ^{2}+\sum_{k=2}^{n}\binom{n}{k}\left( 2\overline{u}\phi+\phi^{2}\right) ^{n-k}\left( 2(\overline{u}+\phi)+\psi\right) ^{k}\psi^{k}~. \label{that} \end{equation} \subsection{The function $V\label{sectionv}$} The function $\widetilde{\mu},$ $\widetilde{\mu}(y)=\mu_{1}(y)/y$ is strictly decreasing on $\mathbf{R}_{+},$ and therefore $\mu_{1}(y)/y\geq\mu _{1}(t^{\gamma}y)/(t^{\gamma}y)$ for $t\geq1.$ Furthermore, the functions $% \eta$ and $\mu_{1}$ are strictly positive and $\mu_{1}$ is strictly increasing. These properties imply that, for $t\geq\tau\geq1,$ $\overline {u}(x,t)+t^{-\gamma}\eta(\left| z\right| )=$ $\mu_{1}(\left| y\right| )+t^{-\gamma}\eta(\left| z\right| )\geq$ $t^{-\gamma}(\mu_{1}(\left| z\right| )+\eta(\left| z\right| ))$ $\geq t^{-\gamma}c_{0}>0,$ where $% c_{0}=\inf_{z>0}(\mu_{1}(z)+\eta(z)).$ Next, since the functions $\mu_{3}$ and $\varphi_{2}$ are bounded and since $3\gamma\geq\varepsilon,$ for $% n\geq4,$ we have that $\left| t^{-\varepsilon}\mu_{3}(\left| y\right| )+t^{-3\gamma}\varphi_{2}(\left| z\right| )\right| <\mathrm{const.~}% t^{-\varepsilon},$ and as a consequence $(\overline{u}+\phi)$ and $(2% \overline{u}+\phi)$ are positive functions of $x$ for all fixed $t\geq \tau_{0},$ if $\tau_{0}$ large enough. \begin{proposition} \label{propvodd}For $n$ odd, $n\geq 5,$ there exists $\tau _{0}\geq 1,$ such that for all $t\geq \tau _{0}$ the function $\widehat{V}$ is positive. \end{proposition} \begin{proof} The function $(2\overline{u}\phi +\phi ^{2})^{n-1}$ is positive, for $n$ odd. \end{proof} \medskip As a consequence, for $n$ odd, equation (\ref{equpsi1}) is of the form indicated in Section 2, with $V=\widehat{V}$ and $T=\widehat{T}.$ The rest of this section treats the case of $n$ even, which, as indicated in the introduction, is slightly more delicate. It can be skipped in a first reading or if the reader is only interested in the case of $n$ odd. So let $n$ be even. The idea is to split $\widehat{V}$ into its positive part $V=$ $\widehat{V}_{+}$ and its negative part $V_{1}=\widehat{V}_{-}~$, and to show that $\widehat{V}_{-}$ is small enough so that it can be treated together with the nonlinear term. Consider the function $\phi $ defined in (% \ref{defphi}). The problem is that $\phi $ becomes negative for large values of $x$, and that therefore $\widehat{V}$ becomes negative for large values of $x.$ To understand why $\phi $ becomes negative, we note that the leading order term $t^{-\gamma }\lambda z^{-\delta }$ in the large $z$ asymptotics of $t^{-\gamma }\eta (z)$ is compensated by the leading order term $% -t^{-\varepsilon }\lambda y^{-\delta }$ in the large $y$ asymptotics of $% t^{-\varepsilon }\mu _{3}(y).$ The leading order of $\phi $ at $x$ large is therefore given by the second order term in the large $z$ asymptotics of $% \eta $ and the leading term in the large $z$ asymptotics of $\varphi _{2}.$ The first of these terms is proportional to $t^{-\gamma }z^{-\delta ^{\prime }},$ and the second one is proportional to $t^{-3\gamma }z^{2-\delta ^{\prime }}\equiv t^{-\gamma }y^{2}z^{-\delta ^{\prime }}.$ The corresponding proportionality constants $\lambda _{\infty }$ and $\lambda ^{\prime }$ can be computed for $n>5$ and turn out to be negative. For $% n=4,5 $ these constants can not be obtained from asymptotic expansions, but numerical results show that they are in fact also negative in these cases. We do not need a proof of this numerical fact, because the following proposition is also correct for positive $\widehat{V}$. \begin{proposition} \label{propV}For $n$ even, $n\geq4,$ there exists $\tau_{1}\geq1$, such that the function $V_{1}$, satisfies for all $t\geq\tau_{1}$ the bound \begin{equation} \sup_{x\in\mathbf{R}}|V_{1}(x,t)|\leq\mathrm{const.}~t^{-\gamma(n-1)(\delta ^{\prime}+1)}~. \label{v1bound} \end{equation} \end{proposition} \begin{proof} The idea is to write $\phi $ as the sum of a function $\phi _{0}$ that is positive and a function $\phi _{\infty }$ that absorbs the asymptotic behavior at infinity. Since $\mu _{3}(y)\approx \lambda _{0}y^{2-\delta }$ for $y$ small, with $\lambda _{0}>0,$ there exists $y_{0}>0$ such that $\mu _{3}(|y|)\geq 0,$ for all $|y|\leq y_{0}.$ Let $c>0,$ to be chosen below, and let $\theta $ be the Heaviside step function, i.e., $\theta (x)=1$ for $% x>0,$ and $\theta (x)=0$ for $x<0.$ Then, we define the function $\phi _{\infty }$ by the equation \begin{equation*} \phi _{\infty }(x,t)=-ct^{-\gamma }\theta (\left| y\right| -y_{0})\left| y\right| ^{2}\left| z\right| ^{-\delta ^{\prime }}~, \end{equation*} and we set $\phi _{0}=\phi -\phi _{\infty }.$ In order to prove that $\phi _{0}$ is positive, for $c$ large enough, we write $\phi _{0}=\phi _{0}^{\left( 1\right) }+\phi _{0}^{\left( 2\right) },$ where \begin{equation*} \phi _{0}^{(1)}(x,t)=t^{-\gamma }(\eta (\left| z\right| )-\lambda |z|^{-\delta }\theta (\left| y\right| -y_{0}))+t^{-3\gamma }\varphi _{2}(\left| z\right| )+c~\theta (\left| y\right| -y_{0})t^{-3\gamma }|z|^{2-\delta ^{\prime }}~, \end{equation*} and \begin{equation*} \phi _{0}^{(2)}(x,t)=t^{-\varepsilon }\left( \mu _{3}(\left| y\right| )+\lambda |y|^{-\delta }\theta (\left| y\right| -y_{0})\right) ~. \end{equation*} $\phi _{0}^{(2)}$ is positive for $\left| y\right| >y_{0},$ since in this case $\phi _{0}^{(2)}(x,t)=\mu _{2}(\left| y\right| )>0,$ and $\phi _{0}^{(2)}$ is positive for $\left| y\right| 0$ for all $z\geq 0,$ and all $t\geq \tau ,$ if $\tau $ is sufficiently large, since $\eta >0,$ since $\varphi _{2}$ is bounded, and since $\left| \varphi _{2}(z)\right| <\eta (z)$ for $z$ large enough. Finally, using the asymptotic properties of $\eta $ and $\varphi _{2} $ we see that $\phi _{0}^{(1)}>0$ for $\left| z\right| >t^{\gamma }y_{0} $ if $c$ is chosen large enough. We now estimate the function $V_{1}.$ From the definition of $\phi _{\infty } $ we get that \begin{equation*} |\phi _{\infty }(x,t)|\leq \mathrm{const.}\text{ }t^{-\gamma (\delta ^{\prime }+1)}~, \end{equation*} and therefore, since $\phi _{0}$ is positive, we have the lower bound \begin{equation*} \phi (x,t)^{n-1}\geq \overline{c}~t^{-\gamma (n-1)(\delta ^{\prime }+1)}~, \end{equation*} for some constant $\overline{c}<0$, from which (\ref{v1bound}) follows. \end{proof} \subsection{The function $I$} \begin{theorem} \label{thm:Inhomo} Let $I$ be as defined in (\ref{equi}), and let $n\geq4.$ Then, there exists a constant $c_{I}>0,$ such that for all $t\geq1,$% \begin{equation} \int_{-\infty}^{\infty}dx~|I(\sqrt{t}x,t)|\leq c_{I}~t^{-1-4\gamma }~. \label{eqn:thminhomo} \end{equation} \end{theorem} The function $I$ is even, and it is therefore sufficient to bound it for $% x\geq0.$ The strategy of the proof will be to rewrite the function $I$ as a sum of functions of the form $t^{-\sigma}G(y)F(z),$ with $\sigma>0,$ and with $G$ and $F$ functions with appropriate asymptotic behavior at zero and infinity. Each of the terms in the sum can then be estimated with the help of Lemma \ref{lem:estimations} below. In order to keep the notation as simple as possible, we suppress in what follows the arguments of functions whenever there is no risk of confusion. \begin{proposition} For $x\geq0,$ the function $-I$ is of the form \begin{equation} -I=\sum_{p=2}^{n}\sum_{q=0}^{p}A_{p,q}+\sum_{i=2}^{8}A_{i}~, \label{repi} \end{equation} where \begin{align*} A_{2} & =\sum_{p=1}^{n-1}\binom{n-1}{p}t^{-2\gamma(n-1-p)-p% \varepsilon}T_{1}^{n-1-p}T_{2}^{p}\left( t^{-4\gamma}T_{3}\right) ~, \\ A_{3} & =n(t^{-2\gamma}T_{1}+t^{-\varepsilon}T_{2})^{n-1}\left( t^{-\gamma(1+\delta^{\prime})}T_{4}+t^{-2\varepsilon}T_{5}+t^{-6\gamma}T_{6}% \right) ~, \\ A_{4} & =\sum_{p=2}^{n}\binom{n}{p}(t^{-2\gamma}T_{1}+t^{-% \varepsilon}T_{2})^{n-p}(t^{-4\gamma}T_{3}+t^{-\gamma(1+\delta^{% \prime})}T_{4}+t^{-2\varepsilon}T_{5}+t^{-6\gamma}T_{6})^{p}~, \\ A_{5} & =-t^{-1-3\gamma}\left( 3\gamma\varphi_{2}+\alpha z\varphi _{2}^{\prime}\right) ~, \\ A_{6} & =-t^{-2n\gamma-2\gamma}nT_{1}^{n-1}2z^{2-\delta}\eta\left( \lambda_{0}-\mu_{3}y^{\delta-2}\right) ~, \\ A_{7} & =t^{-2n\gamma+2\gamma-\varepsilon}nT_{1}^{n-1}(2\mu_{1}\mu _{3}-2\kappa\lambda_{0}y^{3-\delta})-t^{-2n\gamma+2\gamma-\varepsilon }n(2\kappa\lambda z^{1-\delta})^{n-1}(2\mu_{1}\mu_{3}-2\kappa\lambda _{0}y^{3-\delta})~, \\ A_{8} & =t^{-2n\gamma+\gamma}nT_{1}^{n-1}2\eta((\mu_{1}-\kappa y)-\kappa _{3}y^{3})-t^{-2n\gamma+\gamma}n(2\kappa\lambda z^{1-\delta})^{n-1}2\lambda z^{-\delta}(\mu_{1}-\kappa y-\kappa_{3}y^{3})~, \\ A_{p,q} & =\binom{n}{p}\binom{p}{q}\left( R_{2}^{p,q}-R_{1}^{p,q}\right) ~, \end{align*} where \begin{align*} R_{1}^{p,q} & =t^{-2n\gamma+2p\gamma-p\varepsilon}(2\kappa\lambda z^{1-\delta})^{n-p}(2(\mu_{1}-\kappa y)\lambda y^{-\delta})^{p-q}(2\mu_{1}\mu_{3})^{q}~, \\ R_{2}^{p,q} & =t^{-2n\gamma+2p\gamma-p\varepsilon}T_{1}^{n-p}(2(\mu _{1}-\kappa y)y^{-\delta}z^{\delta}\eta)^{p-q}(2\mu_{1}\mu_{3})^{q}~, \end{align*} and where \begin{align} T_{1}(z) & =2\kappa z\eta(z)+\eta(z)^{2}~, \label{t1} \\ T_{2}(y,z) & =2(\mu_{1}(y)-\kappa y)y^{-\delta}z^{\delta}\eta(z)+2\mu _{1}(y)\mu_{3}(y)~, \label{t2} \\ T_{3}(y,z) & =\left( 2\kappa z+2\eta(z)\right) \varphi_{2}(z)+2\mu _{3}(y)y^{\delta-2}z^{2-\delta}\eta(z)~, \label{t3} \\ T_{4}(y,z) & =2(\mu_{1}(y)-\kappa y)y^{2-\delta^{\prime}}z^{\delta^{\prime }-2}\varphi_{2}(z)~, \label{t4} \\ T_{5}(y) & =\mu_{3}(y)^{2}~, \label{t5} \\ T_{6}(y,z) & =2\mu_{3}(y)y^{\delta-2}z^{2-\delta}\varphi_{2}(z)+\varphi _{2}(z)^{2}~. \label{t6} \end{align} \end{proposition} \begin{proof} In terms of the functions (\ref{t1})--(\ref{t6}) we get that, for $x>0,$ \begin{equation*} 2\overline{u}\phi +\phi ^{2}=t^{-2\gamma }T_{1}+t^{-\varepsilon }T_{2}+t^{-4\gamma }T_{3}+t^{-\gamma (1+\delta ^{\prime })}T_{4}+t^{-2\varepsilon }T_{5}+t^{-6\gamma }T_{6}~, \end{equation*} and therefore $(2\overline{u}\phi +\phi ^{2})^{n}=B_{1}+\sum_{i=2}^{4}A_{i}$% , where \begin{equation*} B_{1}=(t^{-2\gamma }T_{1}+t^{-\varepsilon }T_{2})^{n}+n(t^{-2\gamma }T_{1})^{n-1}(t^{-4\gamma }T_{3})~, \end{equation*} and where $A_{2},$ $A_{3}$ and $A_{4}$ are as defined above. Since $-I=\dot{% \phi}-\phi _{1}^{\prime \prime }+\left( 2\overline{u}\phi +\phi ^{2}\right) ^{n},$ it remains to be shown that \begin{equation*} B_{1}-\phi _{1}^{\prime \prime }+\dot{\phi}=\sum_{p=2}^{n}% \sum_{q=0}^{p}A_{p,q}+\sum_{i=5}^{8}A_{i}~. \end{equation*} Using the differential equations for $\mu _{2},$ $\eta $ and $\varphi _{2},$ we find that \begin{equation*} B_{1}-\phi _{1}^{\prime \prime }+\dot{\phi}=R_{1}+R_{2}+S_{3}+A_{5}~, \end{equation*} where $A_{5}$ as defined above, where \begin{align*} R_{1}& =-t^{-n\varepsilon }\left( -(2\kappa \lambda y^{1-\delta })^{n}+(2\mu _{1}(\lambda y^{-\delta }+\mu _{3}))^{n}\right) ~, \\ R_{2}& =(t^{-2\gamma }T_{1}+t^{-\varepsilon }T_{2})^{n}-(t^{-2\gamma }T_{1})^{n}~, \end{align*} and where \begin{align*} S_{3}& =-t^{-2n\gamma -2\gamma }\left[ -n(2\kappa \lambda )^{n}(\frac{\kappa _{3}}{\kappa }+\frac{\lambda _{0}}{\lambda })z^{-\delta }+n(T_{1})^{n-1}\left( (\varphi _{2}+\lambda _{0}z^{2-\delta })(2\kappa z+2\eta )+2\kappa _{3}z^{3}\eta \right) \right] \\ & \hspace{3cm}+n(t^{-2\gamma }T_{1})^{n-1}(t^{-4\gamma }T_{3})~. \end{align*} The functions $R_{1}$ and $R_{2}$ can be further decomposed as follows \begin{align*} R_{1}& =S_{1}-\sum_{p=2}^{n}\sum_{q=0}^{p}\binom{n}{p}\binom{p}{q}% R_{1}^{p,q}~, \\ R_{2}& =S_{2}+\sum_{p=2}^{n}\sum_{q=0}^{p}\binom{n}{p}\binom{p}{q}% R_{2}^{p,q}~, \end{align*} where $R_{1}^{p,q}$ and $R_{2}^{p,q}$ are as defined above, and where \begin{align*} S_{1}& =-nt^{-2\gamma (n-1)-\varepsilon }(2\kappa \lambda z^{1-\delta })^{n-1}(2(\mu _{1}-\kappa y)\lambda y^{-\delta }+2\mu _{1}\mu _{3})~, \\ S_{2}& =nt^{-2\gamma (n-1)-\varepsilon }T_{1}^{n-1}T_{2}~. \end{align*} It remains to be shown that \begin{equation*} \sum_{i=1}^{3}S_{i}=\sum_{i=6}^{8}A_{i}~, \end{equation*} but this follows using the definitions. \end{proof} \subsubsection{Proof of Theorem \ref{thm:Inhomo}} In order to characterize the behavior of a function near zero and infinity we introduce the following family of vector spaces. \begin{definition} Let $p$ and $q$ be two real numbers with $p+q\geq0.$ Then, we define $% \mathcal{V}(p,q)$ to be the vector space of continuous functions $F$ from $% \mathbf{R}_{+}$ to $\mathbf{R,}$ for which the norm \begin{equation*} \left\| F\right\| _{p,q}=\sup_{x\geq0}\left| F(x)\right| \left( x^{-p}+x^{q}\right) \end{equation*} is finite. \end{definition} Note that, if a function is in $\mathcal{V}(p,q)$, then it is also in $% \mathcal{V}(p^{\prime},q^{\prime})$ for any pair of numbers $p^{\prime}\leq p,q^{\prime}\leq q$ for which$\ p^{\prime}+q^{\prime}\geq0.$ Furthermore, if $F_{1}$ is in $\mathcal{V}(p_{1},q_{1})$, and $F_{2}$ is in $\mathcal{V}% (p_{1},q_{2})$, then the product $F_{1}F_{2}$ is in $\mathcal{V}% (p_{1}+p_{2},q_{1}+q_{2})$. The following Lemma provides the tool that we use to estimate the terms on the right hand side of (\ref{repi}). \begin{lemma} \label{lem:estimations} \label{lem:bornes} Let $F\in\mathcal{V}(F_{0},F_{1}) $ and $G\in\mathcal{V}(G_{0},G_{1})$, and assume that \begin{align} 1-G_{1} & -F_{0}~, \label{cb2} \end{align} and that \begin{equation} F_{1}\neq1+G_{0}~. \label{cb3} \end{equation} Then, there is a constant $C>0,$ such that for all $t\geq1,$ \begin{equation} \int_{0}^{\infty}\left| G(x)F(t^{\gamma}x)\right| ~dx\leq Ct^{-\xi }~, \label{decay} \end{equation} where \begin{equation} \xi=\gamma\cdot\min\{F_{1},1+G_{0}\}~. \label{cb4} \end{equation} \end{lemma} \begin{proof} From (\ref{cb3}) it follows that either $F_{1}<1+G_{0}$ or $F_{1}>1+G_{0}.$ In the first case we get using (\ref{cb1}) that $1-G_{1}<$ $F_{1}<$ $% 1+G_{0}, $ and therefore \begin{align*} \int_{0}^{\infty}\left| G(x)F(t^{\gamma}x)\right| ~dx & \leq\left( \sup_{x\geq0}x^{F_{1}}\left| F(t^{\gamma}x)\right| \right) \int_{0}^{\infty}% \frac{1}{x^{F_{1}}}\left| G(x)\right| ~dx \\ & \leq\mathrm{const.}~t^{-\gamma F_{1}}~, \end{align*} and in the second case we get using (\ref{cb2}) that $-F_{0}<1+G_{0}0$ we have the bounds \begin{align} T_{2}(y,z) & \leq\widehat{T_{2}}(y)\equiv2\left| \mu_{1}-\kappa y\right| y^{-\delta}\left( \sup_{z>0}\left| z^{\delta}\eta(z)\right| \right) +2\left| \mu_{1}\mu_{3}\right| ~, \notag \\ T_{3}(y,z) & \leq\widehat{T_{3}}(z)\equiv\left| \left( 2\kappa z+2\eta\right) \varphi_{2}\right| +2\left( \sup_{y>0}\left| \mu _{3}(y)y^{\delta-2}\right| \right) z^{2-\delta}\eta~, \notag \\ T_{4}(y,z) & \leq T_{4,1}(y)\cdot T_{4,2}(z)\equiv\left( 2\left| \mu _{1}(y)-\kappa y\right| y^{2-\delta^{\prime}}\right) \cdot\left( z^{\delta^{\prime}-2}\left| \varphi_{2}(z)\right| \right) ~, \label{inequalities} \\ T_{6}(y,z) & \leq\widehat{T_{6}}(z)\equiv2\left( \sup_{y>0}\left| \mu _{3}(y)y^{\delta-2}\right| \right) z^{2-\delta}\left| \varphi_{2}\right| +\left| \varphi_{2}\right| ^{2}~, \notag \end{align} and $T_{1}\in\mathcal{V}(0,\delta-1)$, $\widehat{T_{2}}\in\mathcal{V}% (3-\delta,\delta-1)$, $\widehat{T_{3}}\in\mathcal{V}(2-\delta,\delta^{\prime }-3)$, $T_{4,1}\in\mathcal{V}(5-\delta^{\prime},\delta^{\prime}-3)$, $% T_{4,2}\in\mathcal{V}(\delta^{\prime}-\delta,0)$, $T_{5}\in\mathcal{V}% (4-2\delta,2\delta)$, $\widehat{T_{6}}\in\mathcal{V}(4-2\delta,\delta^{% \prime }+\delta-4)$. \end{proposition} \begin{proof} The inequalities (\ref{inequalities}) follow by using the triangle inequality and the asymptotic properties of the functions $\mu_{1}$, $% \mu_{2} $, $\eta$ and $\varphi_{1}$. \end{proof} \paragraph{Bound on the function $A_{2}$.} We have the bound \begin{equation*} |A_{2}|\leq\mathrm{const.}\sum_{p=1}^{n-1}t^{-\sigma}\widehat{T_{2}}% ^{p}~\left( T_{1}^{n-1-p}\widehat{T_{3}}\right) ~, \end{equation*} where $\sigma=1+\gamma+3p\varepsilon\gamma.$ The function $G=\widehat{T_{2}}% ^{p}$ is in $\mathcal{V}((3-\delta)p,3\varepsilon p)$, and the function $% F=T_{1}^{n-1-p}\widehat{T_{3}}$ is in $\mathcal{V}(2-\delta,$ $3\varepsilon (n-1-p)-3+\delta^{\prime})$. Since $\delta^{\prime}>1,$ the inequalities (% \ref{cb1}) and (\ref{cb2}) are satisfied and, since $\delta^{\prime}<5$ for $% n\geq3,$% \begin{equation*} \xi/\gamma=\left\{ \begin{array}{ll} \delta^{\prime}-3\varepsilon p & \text{if}\;p\geq2 \\ 3-3\varepsilon p & \text{if}\;p=1 \end{array} \right. \geq3-3\varepsilon p~. \end{equation*} Therefore, $\sigma+\xi\geq$ $1+\gamma+3p\varepsilon\gamma+(3-3p\varepsilon )\gamma=$ $1+4\gamma\ $as required. \paragraph{Bound on the function $A_{3}$.} We have that $A_{3}=t^{-\gamma(1+\delta^{\prime})}B_{3,4}+t^{-2\varepsilon }B_{3,5}+t^{-6\gamma}B_{3,6},$ where $B_{3,i}=n(t^{-2\gamma}T_{1}+t^{-% \varepsilon}T_{2})^{n-1}T_{i},$ $i=4,\dots,6$. Since $|T_{2}/T_{1}|\leq% \mathrm{const.}~t^{3\gamma\varepsilon}$, and $\varepsilon-3\gamma \varepsilon=2\gamma,$ we have the bound \begin{equation*} t^{-\gamma(1+\delta^{\prime})}|B_{3,4}|\leq\mathrm{const.}\text{ }t^{-\sigma }\left( T_{1}^{n-1}T_{4,2}\right) ~T_{4,1}~, \end{equation*} with $\sigma=2\gamma(n-1)+\gamma(1+\delta^{\prime}).$ The function $% G=T_{4,1} $ is in $\mathcal{V}(5-\delta^{\prime},\delta^{\prime}-3)$ and the function $F=T_{1}^{n-1}T_{4,2}$ is in $\mathcal{V}(\delta^{\prime}-\delta,3)$% . Since $\delta^{\prime}>3$ the inequalities (\ref{cb1}) and (\ref{cb2}) are satisfied and $\xi/\gamma=6-\delta^{\prime}$. Therefore $\sigma+\xi=$ $% 1-3\gamma +\gamma(1+\delta^{\prime})+(6-\delta^{\prime})\gamma=$ $1+4\gamma$ as required. Similarly, we have that \begin{equation*} t^{-2\varepsilon}|B_{3,5}|\leq\mathrm{const.}\text{ }t^{-% \sigma}T_{1}^{n-1}~T_{5}~, \end{equation*} with $\sigma=2\gamma(n-1)+2\varepsilon.$ The function $G=T_{5}$ is in $% \mathcal{V}(4-2\delta,2\delta)$ and the function $F=T_{1}^{n-1}$ is in $% \mathcal{V}(0,3)$. The inequalities (\ref{cb1}) and (\ref{cb2}) are satisfied and $\xi/\gamma=5-2\delta$ $=3-6\varepsilon.$ Therefore, $% \sigma+\xi=$ $1-3\gamma+2\varepsilon+(3-6\varepsilon)\gamma=$ $1+4\gamma$ as required. Finally, \begin{equation*} t^{-6\gamma}|B_{3,6}|\leq\mathrm{const.}\text{ }t^{-\sigma}T_{1}^{n-1}% \widehat{T_{6}}~, \end{equation*} where $\sigma=2(n-1)\gamma+6\gamma.$ The function $G\equiv1$ is in $\mathcal{% V}(0,0)$, and the function $F=T_{1}^{n-1}\widehat{T_{6}}$ is in $\mathcal{V}% (4-2\delta,3(n-1)\varepsilon+\delta^{\prime}+\delta-4)$. The inequalities (% \ref{cb1}) and (\ref{cb2}) are satisfied and $\xi/\gamma=1$. Therefore, $% \sigma+\xi=$ $1-3\gamma+6\gamma+\gamma=$ $1+4\gamma$ as required. \paragraph{Bound on the function $A_{4}$.} Since the functions $T_{3}/T_{1}$ and $T_{6}/T_{1}$ are bounded, $% T_{4}/T_{1}\leq\mathrm{const.}~t^{3\varepsilon\gamma}$ and $T_{5}/T_{1}\leq% \mathrm{const.}~t^{3\varepsilon\gamma}$ we have that \begin{equation*} |A_{4}|\leq\mathrm{const.}\sum_{p=2}^{n}t^{-2\gamma(n+p)}T_{1}^{n}\leq% \mathrm{const.}~t^{-\sigma}T_{1}^{n}~, \end{equation*} where $\sigma=2n\gamma+4\gamma.$ The function $G\equiv1$ is in $\mathcal{V}% (0,0)$, and the function $F=T_{1}^{n}$ is in $\mathcal{V}(0,3n\varepsilon) $% . The inequalities (\ref{cb1}) and (\ref{cb2}) are satisfied and $% \xi/\gamma=1$. Therefore, $\sigma+\xi=$ $\left( 1-\gamma\right) +4\gamma+\gamma=$ $1+4\gamma$ as required. \paragraph{Bound on the function $A_{5}$.} We have the bound \begin{equation} \left| A_{5}\right| \leq t^{-\sigma}\left| 3\gamma\varphi_{2}+\alpha z\varphi_{2}^{\prime}\right| ~, \end{equation} where $\sigma=2n\gamma+4\gamma.$ The function $G\equiv1$ is in $\mathcal{V}% (0,0)$, and the function $F=\left| 3\gamma\varphi_{2}+\alpha z\varphi _{2}^{\prime}\right| $ is in $\mathcal{V}(2-\delta,\delta^{\prime}-2)$. The inequalities (\ref{cb1}) and (\ref{cb2}) are satisfied and $\xi/\gamma=1$. Therefore, $\sigma+\xi=$ $\left( 1-\gamma\right) +4\gamma+\gamma=$ $% 1+4\gamma $ as required. \paragraph{Bound on the function $A_{6}$.} We have the bound \begin{equation*} |A_{6}|\leq\mathrm{const.}\text{ }t^{-\sigma}\left( T_{1}^{n-1}z^{2-\delta }\eta\right) ~|\lambda_{0}-\mu_{3}y^{\delta-2}|~, \end{equation*} where $\sigma=1+\gamma.$ The function $G=|\lambda_{0}-\mu_{3}y^{\delta-2}|$ is in $\mathcal{V}(2,2)$ and the function $F=T_{1}^{n-1}z^{2-\delta}\eta$ is in $\mathcal{V}(2-\delta,1+2\delta).$ The inequalities (\ref{cb1}) and (\ref {cb2}) are satisfied and $\xi/\gamma=3$. Therefore, $\sigma+\xi=$ $% 1+\gamma+3\gamma=$ $1+4\gamma$ as required. \paragraph{Bound on the function $A_{7}$.} We have the bound \begin{equation*} |A_{7}|\leq\mathrm{const.}\text{ }t^{-\sigma}|T_{1}^{n-1}-(2\kappa\lambda z^{1-\delta})^{n-1}|~|2\mu_{1}\mu_{3}-2\kappa\lambda_{0}y^{3-\delta}|~, \end{equation*} where $\sigma=2n\gamma-2\gamma+\varepsilon$. The function $G=|2\mu_{1}\mu _{3}-2\kappa\lambda_{0}y^{3-\delta}|$ is in $\mathcal{V}(5-\delta,\delta-3)$ and the function $F=|T_{1}^{n-1}-(2\kappa\lambda z^{1-\delta})^{n-1}|$ is in $\mathcal{V}(-3,3+\delta^{\prime}-\delta).$ The inequalities (\ref{cb1}) and (\ref{cb2}) are satisfied and $\xi/\gamma=6-\delta$. Therefore, $\sigma+\xi=$ $\left( 1-\gamma\right) -2\gamma+\varepsilon+(6-\delta)\gamma=$ $1+4\gamma$ as required. \paragraph{Bound on the function $A_{8}$.} We have the bound \begin{equation*} |A_{8}|\leq \mathrm{const.}\text{ }t^{-\sigma }|T_{1}^{n-1}\eta -(2\kappa \lambda z^{1-\delta })^{n-1}\lambda z^{-\delta }|~|\mu _{1}-\kappa y-\kappa _{3}y^{3}|~, \end{equation*} where $\sigma =2n\gamma -\gamma .$ The function $G=|\mu _{1}-\kappa y-\kappa _{3}y^{3}|$ is in $\mathcal{V}(5,-3)$ and the function $F=|T_{1}^{n-1}\eta -(2\kappa \lambda z^{1-\delta })^{n-1}\lambda z^{-\delta }|\ $is in $% \mathcal{V}(-3-\delta ,3+\delta ^{\prime }).$ The inequalities (\ref{cb1}) and (\ref{cb2}) are satisfied and $\xi /\gamma =6$. Therefore, $\sigma +\xi = $ $\left( 1-\gamma \right) -\gamma +6\gamma =$ $1+4\gamma $ as required. \paragraph{Bound on the functions $A_{p,q}$.} We have the bound \begin{equation*} |A_{p,q}|\leq\mathrm{const.}~t^{-\sigma}~|(z^{\delta}% \eta)^{p-q}~T_{1}^{n-p}-\lambda^{p-q}(2\kappa\lambda z^{1-\delta})^{n-p}|~\left( \left| 2(\mu_{1}-\kappa y)y^{-\delta}\right| ^{p-q}\left| 2\mu_{1}\mu_{3}\right| ^{q}\right) ~, \end{equation*} where $\sigma=2n\gamma-2p\gamma+p\varepsilon.$ The function $G=\left| 2(\mu_{1}-\kappa y)y^{-\delta}\right| ^{p-q}\left| 2\mu_{1}\mu_{3}\right| ^{q}$ is in $\mathcal{V}(p(3-\delta),3\varepsilon p+q)$ and the function $% F=|(z^{\delta}\eta)^{p-q}~T_{1}^{n-p}-\lambda^{p-q}(2\kappa z^{1-\delta })^{n-p}|$ is in $\mathcal{V}(-3\varepsilon(n-p),2+\delta^{\prime }-3p\varepsilon)$. The inequalities (\ref{cb1}) and (\ref{cb2}) are satisfied, and \begin{equation*} \xi/\gamma=\left\{ \begin{array}{ll} 5-3p\varepsilon & \text{if}\;p=2~, \\ 2+\delta^{\prime}-3p\varepsilon & \text{if}\;p\geq3~. \end{array} \right. \end{equation*} Therefore, $\sigma+\xi=1+4\gamma$, for $p=2$ and $\sigma+\xi=1+\gamma (1+\delta^{\prime})>1+4\gamma$, for $p\geq3,$ as required. \bigskip \noindent This completes the proof of Theorem \ref{thm:Inhomo}. $% \blacksquare $ \subsection{The Map $T$} Equation (\ref{equpsi1}) is of the form (\ref{eqpsi}) if we define the map $% T $ by the equation \begin{equation} T(\psi )=\left\{ \begin{array}{ll} \widehat{T}(\psi ) & \text{for }n\text{ odd}~, \\ \widehat{T}(\psi )+V_{1}\psi ~ & \text{for }n\text{ even}~, \end{array} \right. \label{mapt} \end{equation} with $\widehat{T}$ as defined in (\ref{that}) and $V_{1}$ as defined in Section \ref{sectionv}. Using the definitions, we see that $T$ can be written as, \begin{equation} T(\psi )=\sum_{p=1}^{n}\sum_{q=0}^{p}V_{p,q}~\psi ^{p+q}~, \label{eqn:tpsidef} \end{equation} with \begin{equation} V_{p,q}=\left\{ \begin{array}{ll} 0 & \text{for }(p,q)=(1,0)\text{ and }n\text{ odd,} \\ V_{1} & \text{for }(p,q)=(1,0)\text{ and }n\text{ even,} \\ \binom{n}{p}\binom{p}{q}(2\overline{u}\phi +\phi ^{2})^{n-p}(2\overline{u}% +2\phi )^{p-q} & \text{for }p+q\geq 2~. \end{array} \right. \label{vpq} \end{equation} \begin{proposition} Let $V_{p,q}$ as in (\ref{vpq}). Then, for all $t\geq1,$ \begin{equation} \sup_{x\in\mathbf{R}}|V_{p,q}(x,t)|\leq\mathrm{const.}\text{ }t^{-e(p,q)}~, \label{eqn:Lemmavmp} \end{equation} where \begin{equation} e(p,q)=\left\{ \begin{array}{ll} \gamma(n-1)(\delta^{\prime}+1) & \text{for }(p,q)=\left( 1,0\right) ~, \\ 2\gamma(n-2)+2\gamma & \text{for}\;(p,q)=(2,0)~, \\ 2\gamma(n-p) & \text{for}\;(p,q)\neq(2,0)\;\text{and}\;p+q\geq2~. \end{array} \right. \label{mupq} \end{equation} \end{proposition} \begin{proof} The case $(p,q)=(1,0)$ follows from (\ref{v1bound}). Let now $(p,q)\neq (1,0).$ Since $\varepsilon -\gamma -\nu \gamma \geq 0,$ for all $\nu ,$ $% 0\leq \nu \leq \delta ,$ we find that \begin{align*} \sup_{x\in \mathbf{R}}\left| z\right| ^{\nu }~\left| \phi (x,t)\right| & \leq t^{-\gamma }(\sup_{z\in \mathbf{R}_{+}}\left| z^{\nu }~\eta (z)\right| +t^{-(\varepsilon -\gamma -\nu \gamma )}\sup_{y\in \mathbf{R}_{+}}\left| y^{\nu }~\mu _{3}(y)\right| +t^{-2\gamma }\sup_{z\in \mathbf{R}_{+}}\left| z^{\nu }~\varphi _{2}(z)\right| ) \\ & \leq \mathrm{const.}\text{ }t^{-\gamma }~. \end{align*} Furthermore, since $\mu _{1}(y)=\mathcal{O}(y)$ near $y=0,$% \begin{equation*} \left| 2\overline{u}\phi +\phi ^{2}\right| \leq t^{-\gamma }~2\frac{\mu _{1}(\left| y\right| )}{\left| y\right| }~\left| z\phi \right| +\left| \phi \right| ^{2}\leq \mathrm{const.}\text{ }t^{-2\gamma }~. \end{equation*} Since the function $\left| \overline{u}+\phi \right| $ is bounded, it follows that $|V_{p,q}(x,t)|\leq \mathrm{const.}$ $t^{-e(p,q)},$ with $% e(p,q)=2\gamma (n-p).$ For $(p,q)=(2,0)$ we improve this bound using additional properties of the function $\overline{u}+\phi .$ Namely, since $% 2/(n-2)\leq \delta -1,$ we have that \begin{align*} |V_{2,0}(x,t)|& \leq \mathrm{const.}\sup_{x\in \mathbf{R}}\left| (2\overline{% u}\phi +\phi ^{2})^{n-2}(\overline{u}+\phi )^{2}\right| \\ & \leq \mathrm{const.}\sup_{x\in \mathbf{R}}\left| (2t^{-\gamma }\frac{% \overline{u}}{y}z\phi +\phi ^{2})^{n-2}(t^{-\gamma }z\frac{\overline{u}}{y}% +\phi )^{2}\right| \\ & \leq \mathrm{const.}~\sup_{x\in \mathbf{R}}\left| 2t^{-\gamma }\frac{% \overline{u}}{y}z^{1+2/(n-2)}\phi +\left( z^{2/(n-2)}\phi \right) ~\phi \right| ^{n-2}~\left| t^{-\gamma }\frac{\overline{u}}{y}\right| ^{2} \\ & \hspace{3cm}+\mathrm{const.}~\sup_{x\in \mathbf{R}}\left| 2t^{-\gamma }% \frac{\overline{u}}{y}z\phi +\phi ^{2}\right| ^{n-2}~\left( 2t^{-\gamma }\left| \frac{\overline{u}}{y}\right| \left| z\phi \right| +\left| \phi \right| ^{2}\right) \\ & \leq \mathrm{const.}~t^{-2(n-2)\gamma -2\gamma }~. \end{align*} \end{proof} \section{Proof of the main result} For functions $f$ in $\mathcal{J}=L_{1}(\mathbf{R)}\cap L_{\infty }(\mathbf{% R)}$ we use the norms $\left\| f\right\| _{1}=\int |f(x)|~dx$, $\left\| f\right\| _{\infty }=\sup_{x\in \mathbf{R}}|f(x)|$ and $\left\| f\right\| =\left\| f\right\| _{1}+\left\| f\right\| _{\infty }$, and we denote by $% \mathcal{B}$ the Banach space of functions $\varphi $ in $L_{\infty }([1,\infty ))\times \mathcal{J}$ for which the norm $\left\| ~~\right\| _{% \mathcal{B}}$, \begin{equation*} \left\| \varphi \right\| _{\mathcal{B}}=\sup_{t\geq 1}t^{4\gamma }\left\| \varphi (\sqrt{t}~.~,t)\right\| ~, \end{equation*} is finite. Let $\tau _{0}$ as in Proposition \ref{propvodd} and $\tau _{1}$ as in Proposition \ref{propV}, and consider, for fixed $\tau >\max \{\tau _{0},\tau _{1}\},$ functions $\psi $ of the form \begin{equation*} \psi (x,t)=\tau ^{-4\gamma }\varphi (x/\sqrt{\tau },t/\tau )~, \end{equation*} with $\varphi \in \mathcal{B}.$ Let $K$ be the fundamental solution of the differential operator $\partial _{t}-\partial _{x}^{2}-\tau V(\sqrt{\tau }% x,\tau t),$ and let, for given $\nu \in \mathcal{J}$, the map $\mathcal{R}$ be defined by the equation \begin{equation*} \mathcal{R}(\varphi )(x,t)=\varphi _{0,1}(x,t)+\varphi _{0,2}(x,t)+\mathcal{N% }(\varphi )(x,t)~, \end{equation*} where \begin{align*} \varphi _{0,1}(x,t)& =\int_{\mathbf{R}}K(x,t;y,1)~\nu (y)~dy~, \\ \varphi _{0,2}(x,t)& =\tau ^{4\gamma }\tau \int_{1}^{t}ds\int_{\mathbf{R}% }dy~K(x,t;y,s)~I(\sqrt{\tau }y,\tau s)~, \end{align*} and where \begin{equation*} \mathcal{N}(\varphi )(x,t)=\sum_{p=1}^{n}\sum_{q=0}^{p}\mathcal{N}% _{p,q}(\varphi )(x,t)~, \end{equation*} with \begin{equation*} \mathcal{N}_{p,q}(\varphi )(x,t)=\tau ^{4\gamma }\tau \int_{1}^{t}ds\int_{% \mathbf{R}}dy~K(x,t;y,s)~V_{p,q}(\sqrt{\tau }y,\tau s)~\tau ^{-4\gamma (p+q)}\varphi (y,s)^{p+q}~. \end{equation*} The integral equation $\varphi =\mathcal{R}(\varphi )$ is equivalent to the differential equation (\ref{eqpsi}) with initial condition $\psi _{0}(x)=\psi (x,\tau )=\tau ^{-4\gamma }\nu (x/\sqrt{\tau }).$ We note that, since the function $V$ is positive, the kernel $K$ is bounded pointwise by the fundamental solution $K_{0}$ of the heat equation, \begin{equation} K_{0}(x,t;y,s)=\frac{1}{\sqrt{4\pi }}\frac{1}{\sqrt{t-s}}\exp \left( -\frac{1% }{4}\frac{(x-y)^{2}}{(t-s)}\right) ~. \end{equation} The following Proposition makes Theorem \ref{main} precise. \begin{proposition} \label{maint}Let $\beta \geq \max \left\{ 1,~3~c_{I}\int_{0}^{1}\left( 1+% \frac{1}{\sqrt{1-s}}\right) \frac{ds}{s^{1/2+4\gamma }}\right\} ,$ with $% c_{I}$ as defined in (\ref{eqn:thminhomo}), and let $\tau $ be sufficiently large. Then, for all $\nu \in \mathcal{J}$ with $\left\| \nu \right\| <\beta /6$, the equation $\varphi =\mathcal{R}(\varphi )$ has a unique solution $% \varphi ^{\ast }$ in the ball $\mathcal{U}(\beta )=\{\varphi \in \mathcal{B}% | $ $\left\| \varphi \right\| _{\mathcal{B}}<\beta \}.$ \end{proposition} \begin{proof} Since $4\gamma <1/2,$ the solution of the integral equation will be dominated by $\varphi _{0,2},$ and, as we will see, $\beta $ has been chosen such that $\left\| \varphi _{0,2}\right\| _{\mathcal{B}}\leq \beta /3.$ The idea is therefore to show that, if $\tau $ is large enough to make the nonlinear part of the map $\mathcal{R}$ small, and if $\left\| \nu \right\| <\beta /6$, then the map $\mathcal{R}$ contracts the ball $\mathcal{U}(\beta )$ into itself, which by the contraction mapping principle implies the theorem. We first show that $\mathcal{R}$ maps the ball $\mathcal{U}(\beta )$ into itself. For the contribution coming from the initial condition we have \begin{equation*} \left\| \varphi _{0,1}(\sqrt{t}~.~,t)\right\| \leq \frac{2}{\sqrt{t}}\left\| \nu \right\| ~, \end{equation*} and therefore \begin{equation*} \left\| \varphi _{0,1}\right\| _{\mathcal{B}}\leq 2\left\| \nu \right\| <\beta /3~. \end{equation*} We next estimate the norm $\left\| \varphi _{0,2}\right\| _{\mathcal{B}}.$ Let $c(t,s)=\frac{1}{\sqrt{t}}+\frac{1}{\sqrt{t-s}}.$ Then, \begin{align*} \left\| \varphi _{0,2}(\sqrt{t}~.~,t)\right\| & \leq \tau ^{4\gamma }\tau \int_{1}^{t}ds~c(t,s)\int_{\mathbf{R}}dy~\left| I(\sqrt{\tau }y,\tau s)\right| \\ & =\tau ^{4\gamma }\tau \int_{1}^{t}\sqrt{s}~c(t,s)~ds~\int_{\mathbf{R}% }dx~\left| I(\sqrt{\tau s}x,\tau s)\right| \\ & \leq c_{I}~\tau ^{4\gamma }\tau \int_{1}^{t}\sqrt{s}~c(t,s)~ds~\left( \tau s\right) ^{-(1+4\gamma )} \\ & \leq c_{I}~t^{-4\gamma }\int_{0}^{1}c(1,s)~\frac{ds}{s^{1/2+4\gamma }} \\ & \leq \frac{\beta }{3}t^{-4\gamma }~, \end{align*} and therefore \begin{equation*} \left\| \varphi _{0,2}\right\| _{\mathcal{B}}<\beta /3~. \end{equation*} It remains to be shown that the nonlinearity is also bounded by $\beta /3,$ for $\tau $ large enough. For $\varphi \in \mathcal{U}(\beta )$ we have, \begin{equation} \left\| \mathcal{N}(\varphi )(\sqrt{t}~.~,t)\right\| \leq \mathrm{const.}% ~\tau ^{4\gamma }\tau \int_{1}^{t}c(t,s)~\sqrt{s}~ds\sum_{p=1}^{n}% \sum_{q=0}^{p}(\tau s)^{-e(p,q)}~s^{-4\gamma (p+q)}\tau ^{-4\gamma (p+q)}\left\| \varphi \right\| _{\mathcal{B}}^{p+q}~. \label{normcase} \end{equation} For $(p,q)=(1,0)$ we get, since $\delta _{1}\equiv \gamma (n-1)(\delta ^{\prime }+1)-1>0,$ \begin{align*} \left\| \mathcal{N}_{1,0}(\varphi )(\sqrt{t}~.~,t)\right\| & \leq \mathrm{% const.}~\tau ^{-\delta _{1}}\beta \int_{1}^{t}c(t,s)~\sqrt{s}% ~ds~s^{-1-\delta _{1}-4\gamma } \\ & \leq \mathrm{const.}~\tau ^{-\delta _{1}}\beta ~t^{-4\gamma }\int_{0}^{1}% \frac{c(1,s)}{s^{1/2+4\gamma }}~ds~, \end{align*} and for $(p,q)=(2,0)$ we get \begin{align*} \left\| \mathcal{N}_{2,0}(\varphi )(\sqrt{t}~.~,t)\right\| & \leq \mathrm{% const.}~\tau ^{4\gamma +1-8\gamma -(2\gamma (n-2)+2\gamma )}\beta ^{2}\int_{1}^{t}c(t,s)~\sqrt{s}~ds~s^{-8\gamma -2\gamma -2\gamma (n-2)} \\ & \leq \mathrm{const.}~\tau ^{-\gamma }\beta ^{2}~t^{-4\gamma }\int_{0}^{1}% \frac{c(1,s)}{s^{1/2+4\gamma }}~ds~, \end{align*} and for the other cases we have \begin{align*} \left\| \mathcal{N}_{p,q}(\varphi )(\sqrt{t}~.~,t)\right\| & \leq \mathrm{% const.}~\tau ^{4\gamma +1-2\gamma (n-p)-4\gamma p-4\gamma q}\beta ^{2n}\int_{1}^{t}c(t,s)~\sqrt{s}~ds~s^{-2\gamma (n-p)-4\gamma p-4\gamma q} \\ & \leq \mathrm{const.}~\tau ^{-\gamma }\beta ^{2n}~t^{-4\gamma }\int_{0}^{1}% \frac{c(1,s)}{s^{1/2+4\gamma }}~ds~, \end{align*} and therefore $\left\| \mathcal{N}(\varphi )\right\| _{\mathcal{B}}\leq \beta /3$ if $\tau $ is large enough. Using the triangle inequality we get that $\left\| \mathcal{R}(\varphi )\right\| _{\mathcal{B}}\leq \beta ,$ which proves that $\mathcal{R}\left( \mathcal{U}(\beta )\right) \subset \mathcal{U}(\beta )$ as claimed. We now show that $\mathcal{R}$ is Lipschitz. Let $\varphi _{1}$ and $\varphi _{2}$ be in $\mathcal{U}(\beta ).$ We have \begin{align*} \left\| \mathcal{N}(\varphi _{1})(\sqrt{t}~.~,t)-\mathcal{N}(\varphi _{2})(% \sqrt{t}~.~,t)\right\| & \leq \mathrm{const.}~\tau ^{4\gamma }\tau \int_{1}^{t}ds~c(t,s)~\sqrt{s}\cdot \\ & \sum_{p=1}^{n}\sum_{q=0}^{p}(\tau s)^{-e(p,q)}~s^{-4\gamma (p+q)}\tau ^{-4\gamma (p+q)}~\beta ^{p+q-1}~\left\| \varphi _{1}-\varphi _{2}\right\| _{% \mathcal{B}}~, \end{align*} and therefore we get, using the same estimates as for (\ref{normcase}), that \begin{equation*} \left\| \mathcal{R}(\varphi _{1})-\mathcal{R}(\varphi _{2})\right\| _{% \mathcal{B}}=\left\| \mathcal{N}(\varphi _{1})-\mathcal{N}(\varphi _{2})\right\| _{\mathcal{B}}\leq \frac{1}{2}\left\| \varphi _{1}-\varphi _{2}\right\| _{\mathcal{B}}~, \end{equation*} provided $\tau $ is large enough. This completes the proof of Theorem \ref {maint}. \end{proof} \section{Appendix} \subsection{Proof of Proposition \ref{theoremeta}} We first prove the existence of a unique positive solution of equation (\ref {eqeta}) satisfying the boundary conditions (\ref{etainfinity}) and (\ref {etazero}). Then, we derive the results on the asymptotic behavior near zero and infinity. \subsubsection{Existence of the function $\protect\eta\label{seta777}$} \begin{proposition} Let, for $\rho >0$, $\eta _{\rho }$ be the solution of the initial value problem on $\mathbf{R}_{+}$, \begin{align} \eta ^{\prime \prime }& =(2\kappa z\eta +\eta ^{2})^{n}~, \label{1} \\ \eta ^{\prime }(0)& =-\kappa ~, \notag \\ \eta (0)& =\rho >0~. \notag \end{align} Then, there exists a unique $\bar{\rho}$ such that the function $\eta _{% \overline{\rho }}$ is positive and satisfies $\lim_{x\rightarrow \infty }\eta _{\bar{\rho}}(x)=0$. \end{proposition} \begin{proof} We first prove that $\overline{\rho}$ is unique. Given a function $\eta$ from $\mathbf{R}_{+}$ to $\mathbf{R}$ we define the function $\mathcal{F}% (\eta),$ $\mathcal{F}\left( \eta\right) (z)=(\kappa z\eta+\eta^{2})^{n}$. Assume that there are two values $\rho_{1}>\rho_{2}>0,$ such that the functions $\eta _{1}\equiv\eta_{\rho_{1}}$ and $\eta_{2}\equiv\eta_{% \rho_{2}} $ are positive and satisfy $\lim_{x\rightarrow\infty}\eta_{1}(x)=$ $\lim_{x\rightarrow\infty }\eta_{2}(x)=0.$ We first show that the function $% \eta_{12}=\eta_{1}-\eta_{2}$ is positive for all $x\geq0.$ Namely, if we assume the contrary, then because $\eta_{12}(0)>0$, there must be a first $% x_{0}>0$ such that $\eta_{12}(x_{0})=0.$ Furthermore, if $\eta_{12}(x)>0$ then $\eta_{12}^{\prime\prime }(x)=\mathcal{F}\left( \eta_{1}\right) (x)-% \mathcal{F}\left( \eta _{2}\right) (x)>0,$ and therefore $% \eta_{12}(x_{0})=\rho_{1}-\rho_{2}+\int_{0}^{x_{0}}dx\int_{0}^{x}dy~% \eta_{12}^{\prime\prime}(y)>0,$ a contradiction. Therefore $\eta_{12},$ and as a consequence $\eta_{12}^{\prime\prime},$ are positive for all $x,$ from which it follows that $\lim_{x\rightarrow\infty}\eta_{12}(x)>0,$ in contradiction with $\lim_{x\rightarrow\infty}\eta_{12}(x)=\lim_{x\rightarrow% \infty}\eta _{1}(x)-\lim_{x\rightarrow\infty}\eta_{2}(x)=0.$ To prove the existence of a $\bar{\rho}$ for which $\eta _{\overline{\rho }}$ is positive and for which $\lim_{x\rightarrow \infty }\eta _{\bar{\rho}% }(x)=0 $, we use the so called shooting method. Note that, for any $\rho >0$% , the initial value problem (\ref{1}) has a unique solution $\eta _{\rho },$ and since $\eta _{\rho }^{\prime }(0)=-\kappa ,$ the function $\eta _{\rho }$ is strictly decreasing on $[0,x_{\rho })$ for $x_{\rho }$ small enough. We will show that for small enough $\rho >0$, the graph of $\eta _{\rho }$ intersects the real axis and $\eta _{\rho }$ becomes negative, whereas for $% \rho $ large enough, $\eta _{\rho }$ has a minimum and then diverges to plus infinity. The (unique) point between those two sets is $\bar{\rho}$. Define the two subsets $I_{1}$ and $I_{2}$ of $\mathbf{R}_{+},$ \begin{align*} I_{1}& =\{\rho \in \mathbf{R}_{+}|~\exists ~x_{1},\eta _{\rho }(x_{1})=0% \text{{\ \textrm{and\ }}}\eta _{\rho }(x)>0{\ \mathrm{for\ }}x\in \lbrack 0,x_{1})\}~, \\ I_{2}& =\{\rho \in \mathbf{R}_{+}|~\exists ~x_{2},\eta _{\rho }^{\prime }(x_{2})=0{\ \mathrm{and\ }}\eta _{\rho }^{\prime }(x)<0,\eta _{\rho }(x)>0{% \ \mathrm{for\ }}x\in \lbrack 0,x_{2})\}~. \end{align*} We note that if $\eta _{\rho }^{\prime }(x_{0})=0$ and $\eta _{\rho }(x_{0})>0,$ for some $x_{0}$, then $\eta _{\rho }^{\prime }>0$ on any interval $(x_{0},x)$ on which $\eta _{\rho }$ is defined, and a function $% \eta _{\rho }$ with $\rho \in I_{2}$ can therefore not converge to zero at infinity. Furthermore, since the function $\eta \equiv 0$ is a solution of the differential equation (\ref{1}), it follows, since solutions are unique, that $\eta _{\rho }(x_{0})>0$ if $\eta _{\rho }^{\prime }(x_{0})=0,$ and therefore the intersection of $I_{1}$ with $I_{2}$ is empty. The sets $I_{1}$ and $I_{2}$ are open, by continuity of the solution $\eta _{\rho }$ as a function of the initial data $\rho $. We now show that $I_{1}$ is non empty and bounded, which shows that $\bar{\rho}\equiv \sup I_{1}<\infty .$ This $% \bar{\rho}$ is neither in $I_{1}$ nor in $I_{2}$, and therefore the function $\eta _{\bar{\rho}}$ is at the same time strictly positive and strictly decreasing, and therefore $\lim_{x\rightarrow \infty }\eta _{\bar{\rho}% }(x)=0 $. To prove that $I_{1}$ is non empty, we fix any $\rho _{1}$ positive and choose $x_{0}>0$ small enough such that on $[0,x_{0}]$ the solution $\eta _{1}\equiv \eta _{\rho _{1}}$ exists and is strictly decreasing. Then, $\rho _{1}-\eta _{1}(x_{0})>0.$ Choose now $0<\rho _{2}<\rho _{1}-\eta _{1}(x_{0})$ and let $\eta _{2}\equiv \eta _{\rho _{2}}$ be the corresponding solution. As before, we have that the function $\eta _{12}=\eta _{1}-\eta _{2},$ and its second derivative $\eta _{12}^{\prime \prime },$ are positive on the interval $[0,x_{0}),$ and therefore, since $% \eta _{2}(x_{0})=$ $\rho _{2}+\int_{0}^{x_{0}}dx\int_{0}^{x}dy~\eta _{2}^{\prime \prime }(y)$ $=\rho _{2}+\eta _{1}(x_{0})-\rho _{1}-\int_{0}^{x_{0}}dx\int_{0}^{x}dy~\eta _{12}^{\prime \prime }(y),$ we find that $\eta _{2}(x_{0})<\rho _{2}-\rho _{1}+\eta _{1}(x_{0}).$ Using the definition of $\rho _{2}$ we therefore find that $\eta _{2}(x_{0})<0.$ Therefore $\rho _{2}\in I_{1}$. We now prove that $I_{1}$ is bounded. For $% \rho >0$, let $x_{\rho }$ be the largest value (possibly infinite) such that on $[0,x_{\rho })$ the solution $\eta _{\rho }$ exists and is strictly positive. Then, $\eta _{\rho }^{\prime \prime }=\mathcal{F}\left( \eta _{\rho }\right) $ is positive on $(0,x_{\rho })$ and, therefore $\eta _{\rho }(x)>\rho -\kappa x$ for $x\in (0,x_{\rho })$. As a consequence, if the function $\eta _{\rho }$ exists on $[0,\rho /\kappa ],$ then $x_{\rho }\geq \rho /\kappa .$ Using again that $\eta _{\rho }(x)>\rho -\kappa x$ we then find that $\eta _{\rho }(x)>\rho /2$ for $x\in \lbrack 0,\rho /2\kappa ]$, and therefore $\mathcal{F}\left( \eta _{\rho }\right) >(\rho /2)^{2n}$ on $% [0,\rho /2\kappa ]$, which implies that $\eta _{\rho }^{\prime }(\rho /2)>-\kappa +(\rho /2)^{2n+1},$ which is positive if $\rho >2\kappa ^{1/2n+1}.$ Therefore $\eta ^{\prime }(x)$ must be equal to zero for some $% x<\rho /\kappa .$ Any such $\rho $ therefore belongs to $I_{2}.$ If the function $\eta _{\rho }$ ceases to exist before $x=\rho /\kappa $ it must have been diverging to plus infinity for some $x<\rho /\kappa $ which again implies that $\eta _{\rho }^{\prime }(x)$ must have been equal to zero for some $x<\rho /\kappa ,$ and the corresponding $\rho $ is in $I_{2}.$ \end{proof} \subsubsection{Asymptotic behavior of the function $\protect\eta$} The function $\eta $ is regular at zero, and the coefficients of its Taylor series at zero can be computed recursively. We have $\eta (0)=\eta _{0}>0$ and $\eta ^{\prime }(0)=-\kappa ,$ and therefore we get using the differential equation that $\eta _{2}=\eta ^{\prime \prime }(0)/2=\eta _{0}^{2n}/2,$ $\eta ^{\prime \prime \prime }(0)=0$ and $\eta _{4}=-\eta ^{iv}(0)/4!=\frac{n}{12}\eta _{0}^{2n-2}(\kappa ^{2}-\eta _{0}^{2n+1}).$ The asymptotic behavior of $\eta $ at infinity is obtained as follows. Assuming that $\eta $ behaves like $\lambda /z^{\delta }$ at infinity we get from the differential equation that $\delta $ and $\lambda $ are as defined in (\ref {defdelta007}) and (\ref{deflambda007}), respectively. That this is indeed the correct leading behavior of $\eta $ at infinity can now be proved by using standard techniques based on repeated applications of l'H\^{o}pital's rule. See for example \cite{Brezis}. Since the proof is simple, but lengthy and quite uninteresting, we do not give the details here. Once the leading behavior of $\eta $ at infinity has been established we make the ansatz $\eta (z)=\lambda z^{-\delta }+s(z)$. To leading order we get for the function $s$ the linear equation \begin{equation} s^{\prime \prime }-\frac{n}{\lambda }(2\kappa \lambda )^{n}z^{-2}s=n(2\kappa \lambda )^{n-1}\lambda ^{2}z^{-3-2\delta }~. \label{eqs} \end{equation} There is a certain constant $\lambda _{p},$ such that the function $s_{p},$ $% s_{p}(z)=\lambda _{p}z^{-1-2\delta }$ is a particular solution of equation (% \ref{eqs}). The solutions of the homogeneous equation associated with (\ref {eqs}) are of the form $s_{h}^{\pm }(z)=z^{p_{\pm }}$, and using the definition (\ref{deflambda007}) for $\lambda $ we find that \begin{equation} p_{\pm }=\frac{1}{2}\left( 1\pm \sqrt{1+4n\delta (\delta +1)}\right) ~. \label{eqn:critexp} \end{equation} For $n\geq 6$ we have that $\left| p_{-}\right| >2\delta +1,$ and the asymptotic behavior of $s$ is therefore for $n\geq 6$ of the form $\lambda _{\infty }/z^{2\delta +1},$ with $\lambda _{\infty }=\lambda _{p}$, and of the form $\lambda _{\infty }/z^{\left| p_{-}\right| }$ with some unknown coefficient $\lambda _{\infty }$ for $n\leq 5$. It is tedious, but not difficult, to prove that this is indeed the correct second order behavior of $\eta $ at infinity. We omit the details. \subsection{Proof of Proposition \ref{cmu2}} In order to study equation (\ref{mu2}) with boundary conditions (\ref{bmu2-1}% ) and (\ref{bmu2-2}), we make the ansatz $\mu_{2}(x)=m(x)/x^{\delta}.$ For the function $m$ we get the differential equation \begin{equation} m^{\prime\prime}+(\frac{x}{2}-\frac{2\delta}{x})m^{\prime}+\left( \delta(\delta+1)\frac{1}{x^{2}}-\left( \frac{\delta}{2}-\varepsilon\right) \right) m=\frac{1}{x^{2}}(2\frac{\mu_{1}(x)}{x}m)^{n}~, \label{eqh} \end{equation} and the boundary conditions for $m$ are \begin{align} \lim_{x\rightarrow0}m(x) & =\lambda~, \label{bcm0} \\ \lim_{x\rightarrow\infty}m(x)x^{2\varepsilon-\delta} & =0~. \label{bcm1} \end{align} \subsubsection{Asymptotic behavior of the function $\protect\mu_{2}$} As indicated in Section \ref{smu2}, a solution of equation (\ref{mu2}) that is defined on $\mathbf{R}_{+}$ behaves at infinity either like $% x^{-2\varepsilon }$ or like $\exp (-x^{2}/4)/x^{1-2\varepsilon }.$ The proof is similar to the one in \cite{Brezis}. We omit the details. Given the asymptotic behavior of $\mu _{2}$ at infinity, we find for the function $m$ at infinity either a behavior proportional to $x^{\delta -2\varepsilon }$, or a behavior proportional to $x^{5\varepsilon }\exp (-x^{2}/4).$ Since $% \delta -2\varepsilon >0,$ we find that \begin{equation} \lim_{x\rightarrow \infty }m(x)=0~, \label{bcm3} \end{equation} if and only if the boundary condition (\ref{bcm1}) is satisfied, and we will impose (\ref{bcm3}) from now on. We now discuss the asymptotic behavior of the function $m$ near zero. From equation (\ref{eqh}) we see that $m^{\prime \prime }(0)$ exists if and only if $\delta (\delta +1)m(0)=\left( \kappa m(0)\right) ^{n}$, i.e., if $m(0)=\lambda ,$ and if $m^{\prime }(0)=0.$ We then find, that $m^{\prime \prime }(0)/2=\lambda _{0}$, with $\lambda _{0}$ as defined in (\ref{lambda0}). By taking derivatives of equation (\ref{eqh}) we find that $m^{\prime \prime \prime }(0)=0$, and that $m^{iv}(0)/4!=% \lambda _{1},$ for some constant $\lambda _{1}\neq 0.$ By taking further derivatives, one can recursively compute the Taylor coefficients of a solution $m_{0}$ of equation (\ref{eqh}) that is regular (in fact, analytic) in a neighborhood of zero. The solution $m_{0}$ does however not satisfy the boundary condition (\ref{bcm3}). The solution of (\ref{eqh}) that does satisfy (\ref{bcm3}) is of the form \begin{equation} m(x)=m_{0}(x)+x^{p}m_{1}(x)~, \label{msing} \end{equation} where $p=p_{+}+\delta ,$ with $p_{+}$ as defined in (\ref{eqn:critexp}). Here, $m_{1}(x)=m_{1}(0)+\dots $, with $m_{1}^{\prime }(0)=0,$ and with $% m_{1}(0)$ to be determined. The asymptotic form (\ref{msing}) can be obtained by substituting the ansatz (\ref{msing}) for $m$ into equation (\ref {eqh}). Since $p>7$ we find from (\ref{msing}) that near zero $% m_{0}(x)+x^{p}m_{1}(x)=\lambda +\lambda _{0}x^{2}+\lambda _{1}x^{4}+\dots $ . We omit the details of the proof that the asymptotic behavior is as indicated. \subsubsection{Existence of the function $\protect\mu_{2}$} We now prove the existence of a function $m$ that satisfies equation (\ref {eqh}) with the boundary conditions (\ref{bcm0}) and (\ref{bcm3}). Since the second derivative of the solution $m$ at zero is positive, and since $m$ converges to zero at infinity, there must be a first $\xi\in\mathbf{R}_{+}, $ such that $m^{\prime}(\xi)=0.$ The basic idea is now to use this position $% \xi,$ and the value $\rho$ of $m$ at $\xi$, as parameters in shooting arguments towards zero and infinity. The first shooting argument will allow us to define a curve $c_{0}$ of initial conditions $(\xi,\rho)$ in $\mathbf{R% }_{+}^{2},$ for which the boundary condition at zero is satisfied, and the second shooting argument will allow us to find on this curve an initial condition for which the boundary condition at infinity is satisfied as well. So, let $(\xi ,\rho )$ be an initial condition. Locally, i.e., near $\xi ,$ there exists a solution $m_{\xi ,\rho }$ of equation (\ref{eqh}). By definition, $m_{\xi ,\rho }(\xi )=\rho ,$ $m_{\xi ,\rho }^{\prime }(\xi )=0,$ and therefore we get for the second derivative of $m_{\xi ,\rho }$ at $\xi ,$ \begin{equation*} m_{\xi ,\rho }^{\prime \prime }(\xi )=\omega _{1}(\xi )\rho ^{n}+\omega _{2}(\xi )\rho ~, \end{equation*} where \begin{equation} \omega _{1}(\xi )=\frac{\left( 2\frac{\mu _{1}(\xi )}{\xi }\right) ^{n}}{\xi ^{2}}~, \label{omega1} \end{equation} and \begin{equation} \omega _{2}(\xi )=\frac{n}{2}\varepsilon -\frac{\delta (\delta +1)}{\xi ^{2}}% ~. \label{omega2} \end{equation} For initial conditions such that $\rho =c_{2}(\xi ),$ where \begin{equation} c_{2}(\xi )=\left( \frac{\frac{n}{2}\varepsilon }{\left( 2\frac{\mu _{1}(\xi )}{\xi }\right) ^{n}}\left( \xi _{0}^{2}-\xi ^{2}\right) \right) ^{\varepsilon } \label{c2} \end{equation} and $\xi _{0}=\sqrt{\delta (\delta +1)/\left( \frac{n\varepsilon }{2}\right) },$ we therefore have that $m_{\xi ,\rho }^{\prime \prime }(\xi )=0.$ See Fig. 1 for the graph of the function $c_{2}.$ The function $c_{2}$ has a maximum at the point $\xi _{m}$ that satisfies the equation \begin{equation} \omega _{1}^{\prime }(\xi _{m})c_{2}(\xi _{m})^{n-1}+\omega _{2}^{\prime }(\xi _{m})=0~, \label{defxim} \end{equation} and the line $c_{2}$ divides the set of initial conditions into two subsets, a subset $A$ where $m_{\xi ,\rho }^{\prime \prime }(\xi )<0,$ and a subset $% B $ where $m_{\xi ,\rho }^{\prime \prime }(\xi )>0.$ For initial conditions on $c_{2}$ we can compute $m_{\xi ,c_{2}(\xi )}^{\prime \prime \prime }(\xi ),$% \begin{equation*} m_{\xi ,c_{2}(\xi )}^{\prime \prime \prime }(\xi )=\omega _{1}^{\prime }(\xi )c_{2}(\xi )^{n}+\omega _{2}^{\prime }(\xi )c_{2}(\xi )~. \end{equation*} Comparing with (\ref{defxim}) we find that $m_{\xi _{m},c_{2}(\xi _{m})}^{\prime \prime \prime }(\xi _{m})=0,$ and we have that $m_{\xi ,c_{2}(\xi )}^{\prime \prime \prime }(\xi )<0$ for $0<\xi <\xi _{m}.$ We now construct the line $c_{0}$ for $0<\xi <\xi _{m}.$ \begin{proposition} Fix $\xi,$ $0<\xi<\xi_{m}.$ Then, there exists a unique number $c_{0}(\xi),$ $c_{2}(\xi)>$ $c_{0}(\xi)>\lambda,$ such that $m_{\xi,c_{0}(\xi)}$ is positive and satisfies $\lim_{x\rightarrow0}m_{\xi,c_{0}(\xi)}(x)=\lambda.$ Furthermore, the function $c_{0}$ is continuous. \end{proposition} \begin{proof} The proof is similar to the one in Section \ref{seta777}. Define the two subsets $I_{1}$ and $I_{2}$ of the interval $I=(\lambda ,c_{2}(\xi ))$, \begin{align*} I_{1}& =\{\rho \in I|~\exists ~0<\xi _{1}<\xi ,~m_{\xi ,\rho }(\xi _{1})=\lambda \text{ and }\lambda 0$ for $(s,z)$ in the set $C$ (see Fig. 1), and to integrate the difference of two solutions from their respective initial condition to zero, which leads to a contradiction, since both solutions have to be equal to $\lambda $ at zero. Finally, that $c_{0}$ is a continuous function follows from the continuity of $m_{\xi ,\rho }$ as a function of $\rho $ and $\xi $ using the uniqueness of $c_{0}(\xi ).$ \end{proof} We now prove with a second shooting argument that solutions with initial conditions $(\xi,c_{0}(\xi)),$ with $\xi\approx0,$ become negative somewhere in the interval $(\xi,2),$ and that solutions with initial conditions $% (\xi,c_{0}(\xi))$, with $\xi\approx\xi_{m},$ stay positive and diverge to plus infinity. \begin{proposition} There exists a unique initial condition $(\xi^{\ast},c_{0}(\xi^{\ast}))$ such that the corresponding solution $m_{\xi^{\ast},c_{0}(\xi^{\ast})}$ is positive and satisfies $\lim_{x\rightarrow\infty}m_{\xi^{\ast},c_{0}(\xi^{% \ast})}(x)=0.$ \end{proposition} \begin{proof} Define the two subsets $I_{1}$ and $I_{2}$ of the interval $I=(0,\xi _{m})$, \begin{align*} I_{1}& =\{\xi \in I|~\exists ~\xi _{1}>\xi ,~m_{\xi ,c_{0}(\xi )}(\xi _{1})=0% \text{ and }m_{\xi ,c_{0}(\xi )}(x)>0,\text{ }m_{\xi ,c_{0}(\xi )}^{\prime }(x)<0\text{ for }x\in (\xi ,\xi _{1})\}~, \\ I_{2}& =\{\xi \in I|~\exists ~\xi _{2}>\xi ,~m_{\xi ,c_{0}(\xi )}^{\prime }(\xi _{2})=0\text{ and }m_{\xi ,c_{0}(\xi )}(x)>0,\text{ }m_{\xi ,c_{0}(\xi )}^{\prime }(x)<0\text{ for }x\in (\xi ,\xi _{2})\}~. \end{align*} By definition, the intersection of $I_{1}$ with $I_{2}$ is empty, and the sets $I_{1}$ and $I_{2}$ are open, by continuity of the solution $m_{\xi ,c_{0}(\xi )}$ as a function of the initial data $\xi $. We now show that all $\xi $ sufficiently close to $0$ are in $I_{1},$ and that all $\xi $ sufficiently close to $\xi _{m}$ are in $I_{2.}$ This implies that $\xi ^{\ast }=\sup I_{1}<\xi _{m},$ is neither in $I_{1}$ nor in $I_{2}$, and therefore the function $m_{\xi ^{\ast },c_{0}(\xi \ast )}$ is positive and decreasing for $x>\xi ^{\ast }$ which implies that $\lim_{x\rightarrow \infty }m_{\xi ^{\ast },c_{0}(\xi ^{\ast })}(x)=0.$ So let $(\xi ,c_{0}(\xi ))$ be an initial condition with $0<\xi 0,$ and \begin{equation*} k_{2}=\delta (\delta +1)\exp (-\xi _{0}^{2}/4)\int_{\xi _{m}-x_{1}}^{\xi _{0}}z^{-2\delta }~(\frac{1}{\xi _{0}^{2}}-\frac{1}{z^{2}})~dz~, \end{equation*} and for $x\geq \xi _{0}$ we therefore have the lower bound \begin{equation*} m_{\xi ,c_{0}(\xi )}(x)\geq m_{\xi ,c_{0}(\xi )}(\xi )\left( k_{1}+k_{2}\int_{\xi _{0}}^{\infty }\frac{dy}{p(y)}\right) ~, \end{equation*} and it follows, using again the integral equation (\ref{inteq}), that $% m_{\xi ,c_{0}(\xi )}$ diverges at (or before) infinity, that therefore $% m_{\xi ,c_{0}(\xi )}^{\prime }(x)=0$ for some $x>\xi ,$ which implies that $% \xi \in I_{2},$ provided \begin{equation} k_{1}+k_{2}\int_{\xi _{0}}^{\infty }\frac{dy}{p(y)}>0~. \label{cond} \end{equation} For $x_{1}$ small enough and for $n$ large enough (\ref{cond}) can be verified without too much difficulty. With the help of a computer one can show that (\ref{cond}) is satisfied for the remaining $n\geq 5.$ For $n=4$ (% \ref{cond}) is not satisfied, since the above bounds on $m_{\xi ,c_{0}(\xi )}(\xi _{0})$ and $m_{\xi ,c_{0}(\xi )}^{\prime }(\xi _{0})$ are too weak. Sufficiently good bounds can be obtained by dividing the interval $(\xi ,\xi _{0})$ in two pieces and by integrating lower bounds on each of the subintervals. We omit the details. Finally, uniqueness of $\xi ^{\ast }$ can be proved by integrating the difference of two solutions from $\xi _{0}$ to infinity, which, using the positivity of $\partial _{s}\Omega (s,z),$ leads to a contradiction with the fact that both of the solutions converge to zero at infinity. \end{proof} \subsection{Proof of Proposition \ref{cfi22}} We first proof the existence of a unique solution of equation (\ref{h}) with the boundary conditions (\ref{bh-1}) and (\ref{bh-2}). Then, we derive the results on the asymptotic behavior near zero and infinity. \subsubsection{Existence of the function $\protect\varphi_{2}$} The equation (\ref{h}) for $h$ is linear. We therefore first construct two linearly independent solutions $h_{1}$ and $h_{2}$ for the corresponding homogeneous equation, which we then use to construct, using standard methods, a solution of (\ref{h}) that satisfies the boundary conditions (\ref {bh-1}) and (\ref{bh-2}). The homogeneous equation is \begin{equation} h^{\prime \prime }-q~h=0~, \label{homh} \end{equation} where \begin{equation} q(z)=n\left( 2\kappa z\eta (z)+\eta (z)^{2}\right) ^{n-1}(2\kappa z+2\eta (z))~. \label{defq} \end{equation} Since the equation (\ref{homh}) is linear, the integral equation for $h_{1}$% , \begin{equation} h_{1}(x)=1+\int_{0}^{x}~dy\int_{0}^{y}q(z)~h_{1}(z)~dz~, \label{eqh1} \end{equation} has a positive solution that exists for all $x$ in $\mathbf{R}_{+}.$ By definition, we have near $x=0$ the behavior $h_{1}(x)=1+\mathcal{O}(x^{2}).$ At infinity, the solution $h_{1}$ is asymptotic to a solution of the equation \begin{equation*} h^{\prime \prime }(x)-\frac{n}{\lambda }(2\kappa \lambda )^{n}\frac{1}{x^{2}}% h(x)=0~. \end{equation*} This equation is the same as the homogeneous part of equation (\ref{eqs}), and the leading order behavior of $h_{1}$ at infinity is therefore either proportional to $x^{p_{+}}$ or to $x^{p_{-}},$ with $p_{\pm }$ as defined in (\ref{eqn:critexp}). Since $h_{1}$ is positive, we find using (\ref{eqh1}), that $h_{1}(x)>1$ for all $x$ in $\mathbf{R}_{+},$ and therefore $h_{1}$ is near infinity of the form $h_{1}(x)=d_{1}x^{p_{+}}+\dots ,$ for some constant $d_{1}>0$. A second solution of the homogeneous equation (\ref{homh}% ) is \begin{equation*} h_{2}(x)=h_{1}(x)~\int_{0}^{x}\frac{1}{h_{1}(y)^{2}}~dy~. \end{equation*} Near $x=0$ we have that $h_{2}(x)=x+\dots ,$ and near infinity we find that \begin{equation} h_{2}(x)=h_{1}(x)\left( d-d_{2}x^{1-2p_{+}}+\dots \right) ~, \label{h2infty} \end{equation} where $d=\int_{0}^{\infty }1/h_{1}(y)^{2}~dy$, and $d_{2}=$ $\left( 1/d_{1}\right) ^{2}/\left( 2p_{+}-1\right) .$ We note that $% h_{1}h_{2}^{\prime }-h_{1}^{\prime }h_{2}\equiv 1.$ Therefore, the function $% h_{p}$, \begin{equation*} h_{p}(x)=c_{1}(x)~h_{1}(x)+c_{2}(x)~h_{2}(x)~, \end{equation*} where \begin{align*} c_{1}(x)& =-\int_{0}^{x}h_{2}(y)~f(y)dy~, \\ c_{2}(x)& =\int_{0}^{x}h_{1}(y)~f(y)dy~, \end{align*} and where \begin{equation*} f(x)=-\gamma \eta (x)-\alpha x\eta ^{\prime }(x)+n\left( 2\kappa x\eta (x)+\eta (x)^{2}\right) ^{n-1}2\kappa _{3}x^{3}\eta (x)~, \end{equation*} satisfies equation (\ref{h}). Near zero, the function $f$ is of the form $% f(x)=-\gamma \eta _{0}+\dots $, and therefore, using the behavior of $h_{1}$ and $h_{2}$ near zero, we find that $c_{1}$ is near zero of order $\mathcal{O% }(x^{2})$, and $c_{2}$ is near zero of order $\mathcal{O}(x).$ The function $% h_{p}$ is therefore of order $\mathcal{O}(x^{2})$ near zero. At infinity, the function $f$ is of the form $f(x)=f_{\infty }x^{-\delta }+\dots ,$ where $f_{\infty }=-\gamma \lambda +\alpha \lambda \delta +n(2\kappa \lambda )^{n-1}2\kappa _{3}\lambda ,$ and therefore the function $c_{2}$ is near infinity of the form $c_{2}(x)=d_{1}f_{\infty }~x^{p_{+}+1-\delta }/(p_{+}+1-\delta )+\dots $, and $c_{1}$ is near infinity of the form $% c_{1}(x)=-d~c_{2}(x)+h_{\infty }+d_{1}d_{2}f_{\infty }~x^{2-p_{+}-\delta }/(2-p_{+}-\delta )+\dots $, for some constant $h_{\infty }.$ Using these asymptotic behavior for $c_{1}$, $h_{1}$, $c_{2}$, and $h_{2},$ we find for the function $h_{p}$ near infinity the behavior, \begin{align} h_{p}(x)& =(-d~c_{2}(x)+h_{\infty }+\frac{d_{1}d_{2}f_{\infty }}{% (2-p_{+}-\delta )}x^{2-p_{+}-\delta }+\dots )h_{1}(x)+c_{2}(x)h_{1}(x)\left( d-d_{2}x^{1-2p_{+}}+\dots \right) \notag \\ & =h_{\infty }h_{1}(x)-\frac{d_{1}f_{\infty }}{p_{+}+1-\delta }% d_{1}d_{2}x^{p_{+}+1-\delta }x^{p_{+}}x^{1-2p_{+}}+\frac{d_{1}d_{2}f_{\infty }}{2-p_{+}-\delta }d_{1}x^{2-p_{+}-\delta }x^{p_{+}}+\dots \notag \\ & =h_{\infty }h_{1}(x)+\frac{f_{\infty }}{2p_{+}-1}\left( \frac{-1}{% p_{+}+1-\delta }+\frac{1}{2-p_{+}-\delta }\right) x^{2-\delta }+\dots \notag \\ & =h_{\infty }h_{1}(x)+\lambda _{0}x^{2-\delta }+\dots ~. \label{hpinfty} \end{align} In the last equality we have used the definition (\ref{lambda0}) for $% \lambda _{0}.$ The function $h,$% \begin{equation*} h(z)=h_{p}(z)-h_{\infty }~h_{1}(z)~, \end{equation*} solves the equation (\ref{h}), satisfies the boundary condition (\ref{bh-1}% ), and since, as we show in the next section, the higher order terms in (\ref {hpinfty}) converge to zero at infinity, it also satisfies the boundary condition (\ref{bh-2}). \subsubsection{Asymptotic behavior of the function $\protect\varphi_{2}$} By construction, the leading behavior of $h$ at infinity is $h(z)=\lambda _{0}z^{2-\delta }+\dots $. We therefore make the ansatz $h(z)=\lambda _{0}z^{2-\delta }+k(z)$, and to leading order, we get for the function $k$ the linear equation \begin{equation} k^{\prime \prime }-n\delta (\delta +1)z^{-2}k=c_{k}z^{-\delta ^{\prime }}~, \label{bvarphi2-2} \end{equation} for a certain constant $c_{k}$. The general solution of equation (\ref {bvarphi2-2}) is \begin{equation*} k(z)=\frac{\lambda ^{\prime }}{z^{\delta ^{\prime }-2}}+\mathrm{const.}% ~z^{p_{-}}+\mathrm{const.}~z^{p_{+}}~, \end{equation*} with a certain constant $\lambda ^{\prime }$ and with $p_{+}$, $p_{-}$ as defined in (\ref{eqn:critexp}). Since $\lim_{z\rightarrow \infty }k(z)/z^{2-\delta }=0$ but $p_{+}>2-\delta ,$ the coefficient of the term proportional to $z^{p_{+}}$ must be zero. Therefore, since$\left| p_{-}\right| >\delta ^{\prime }-2$ for all $n,$ the asymptotic behavior of $% k $ is always given by $\lambda ^{\prime }/z^{\delta ^{\prime }-2}.$ We omit the details of the proof that this is indeed the correct second order behavior of $k$ at infinity.\bigskip \bigskip \bigskip \noindent \textbf{Acknowledgment} During this work, A.S. was hosted by the University of Chicago and wishes to thank the Department of Mathematics for its hospitality. \bigskip \begin{thebibliography}{9} \bibitem{Wittwer} A. Schenkel, P. Wittwer \& J. Stubbe: Asymptotics of Solutions in an $A+B\rightarrow C$ Reaction--Diffusion System. \textit{% Physica D}, \textbf{\ 69}, 135 (1993). \bibitem{kupi} J. Bricmont, A. Kupiainen \& G. Lin: Renormalisation Group and Asymptotics of Solutions of Nonlinear Parabolic Equations. \textit{% Communications in Pure and Applied Mathematics}, New York, NY ISSN 0010-3640, \textbf{\ 47}, 893, (1994). \bibitem{Droz} S. Cornell, Z. Koza \& M. Droz: Dynamic multiscaling of the Reaction--Diffusion front for $mA+nB\rightarrow 0$. \textit{Phys. Rev. E}, \textbf{52}, 3500, (1995). \bibitem{Brezis} H. Brezis, L. A. Peletier, D. Terman: A Very Singular Solution of the Heat Equation with Absorption.\emph{\ }\textit{Archive for Rational Mechanics and Analysis, }\textbf{95}, 185--209, (1986). \end{thebibliography} \end{document} ---------------9904151203928 Content-Type: application/octet-stream; name="figure.ps" Content-Transfer-Encoding: base64 Content-Disposition: attachment; filename="figure.ps" GyUtMTIzNDVYQFBKTCBKT0IKQFBKTCBTRVQgUkVTT0xVVElPTiA9IDYwMApAUEpMIEVOVEVS IExBTkdVQUdFID0gUE9TVFNDUklQVCAKJSFQUy1BZG9iZS0zLjAKJSVUaXRsZTogZmlndXJl MDA4LmFpCiUlQ3JlYXRvcjogV2luZG93cyBOVCA0LjAKJSVDcmVhdGlvbkRhdGU6IDE3OjM2 IDQvMTUvMTk5OQolJVBhZ2VzOiAoYXRlbmQpCiUlQm91bmRpbmdCb3g6IDE0IDEzIDU4MSA4 MjkKJSVMYW5ndWFnZUxldmVsOiAyCiUlRG9jdW1lbnROZWVkZWRGb250czogKGF0ZW5kKQol JURvY3VtZW50U3VwcGxpZWRGb250czogKGF0ZW5kKQolJUVuZENvbW1lbnRzCiUlQmVnaW5Q cm9sb2cKDQolJUJlZ2luUmVzb3VyY2U6IHByb2NzZXQgTlRQU09jdDk1DQovTlRQU09jdDk1 IDEwMCBkaWN0IGR1cCBiZWdpbi9iZHtiaW5kIGRlZn1iaW5kIGRlZi9sZHtsb2FkIGRlZn1i 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